sites.math.rutgers.edufq15/244s14/memo1-1.pdf · . . . . . . logarithm functions de nitions: f(x) =...
TRANSCRIPT
![Page 1: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/1.jpg)
. . . . . .
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...... Review of Formulas in Calculus
Fei Qi
Rutgers University
January 21, 2014
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 1 / 16
![Page 2: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/2.jpg)
. . . . . .
Disclaimer
The slides are written exclusively for 244 students. It might not beappropriate to use them in any earlier course.
There may be errors. Use them at your own discretion. Anyone whonotify me with an error will get some award in grade points.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 2 / 16
![Page 3: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/3.jpg)
. . . . . .
Power Function
Definitions: f (x) = xa (a ∈ R)
Derivative:
f ′(x) = (xa)′ =
{axa−1 a ̸= 0
0 a = 0
Antiderivative:∫f (x)dx =
∫xadx =
1
a+ 1xa+1 + C a ̸= −1
ln |x |+ C a = −1
Note: When you perform the integration, you should never forget totake absolute values. However in many cases of the 244 course, youdon’t have to care too much about that.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 3 / 16
![Page 4: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/4.jpg)
. . . . . .
Power Function
Definitions: f (x) = xa (a ∈ R)Derivative:
f ′(x) = (xa)′ =
{axa−1 a ̸= 0
0 a = 0
Antiderivative:∫f (x)dx =
∫xadx =
1
a+ 1xa+1 + C a ̸= −1
ln |x |+ C a = −1
Note: When you perform the integration, you should never forget totake absolute values. However in many cases of the 244 course, youdon’t have to care too much about that.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 3 / 16
![Page 5: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/5.jpg)
. . . . . .
Power Function
Definitions: f (x) = xa (a ∈ R)Derivative:
f ′(x) = (xa)′ =
{axa−1 a ̸= 0
0 a = 0
Antiderivative:∫f (x)dx =
∫xadx =
1
a+ 1xa+1 + C a ̸= −1
ln |x |+ C a = −1
Note: When you perform the integration, you should never forget totake absolute values. However in many cases of the 244 course, youdon’t have to care too much about that.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 3 / 16
![Page 6: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/6.jpg)
. . . . . .
Power Function
Definitions: f (x) = xa (a ∈ R)Derivative:
f ′(x) = (xa)′ =
{axa−1 a ̸= 0
0 a = 0
Antiderivative:∫f (x)dx =
∫xadx =
1
a+ 1xa+1 + C a ̸= −1
ln |x |+ C a = −1
Note: When you perform the integration, you should never forget totake absolute values. However in many cases of the 244 course, youdon’t have to care too much about that.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 3 / 16
![Page 7: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/7.jpg)
. . . . . .
Exponential Functions
Definitions: f (x) = ax (a > 0)
Derivative:f ′(x) = (ax)′ = ax ln a.
How to compute: Take logarithms on both sides and apply thedifferentiation law of composite functions.
Antiderivative: ∫f (x)dx =
∫axdx =
1
ln aax + C
How to compute: Make use of the derivative above.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 4 / 16
![Page 8: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/8.jpg)
. . . . . .
Exponential Functions
Definitions: f (x) = ax (a > 0)
Derivative:f ′(x) = (ax)′ = ax ln a.
How to compute: Take logarithms on both sides and apply thedifferentiation law of composite functions.
Antiderivative: ∫f (x)dx =
∫axdx =
1
ln aax + C
How to compute: Make use of the derivative above.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 4 / 16
![Page 9: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/9.jpg)
. . . . . .
Exponential Functions
Definitions: f (x) = ax (a > 0)
Derivative:f ′(x) = (ax)′ = ax ln a.
How to compute: Take logarithms on both sides and apply thedifferentiation law of composite functions.
Antiderivative: ∫f (x)dx =
∫axdx =
1
ln aax + C
How to compute: Make use of the derivative above.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 4 / 16
![Page 10: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/10.jpg)
. . . . . .
Exponential Functions
Definitions: f (x) = ax (a > 0)
Derivative:f ′(x) = (ax)′ = ax ln a.
How to compute: Take logarithms on both sides and apply thedifferentiation law of composite functions.
Antiderivative: ∫f (x)dx =
∫axdx =
1
ln aax + C
How to compute: Make use of the derivative above.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 4 / 16
![Page 11: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/11.jpg)
. . . . . .
Exponential Functions
Definitions: f (x) = ax (a > 0)
Derivative:f ′(x) = (ax)′ = ax ln a.
How to compute: Take logarithms on both sides and apply thedifferentiation law of composite functions.
Antiderivative: ∫f (x)dx =
∫axdx =
1
ln aax + C
How to compute: Make use of the derivative above.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 4 / 16
![Page 12: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/12.jpg)
. . . . . .
Logarithm Functions
Definitions: f (x) = loga x (a > 0)
Derivative:
f ′(x) = (loga x)′ =
1
x ln a.
How to compute: Strictly speaking you should be using the law forinverse functions. But if you know already that (ln x)′ = 1/x , thenyou can simply make use of the fact loga x = ln x/ ln a.
Antiderivative:∫f (x)dx =
∫loga xdx =
1
ln a(x ln x − x) + C
How to compute: Use integration by parts to solve the special casethat a = e, then again use loga x = ln x/ ln a.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 5 / 16
![Page 13: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/13.jpg)
. . . . . .
Logarithm Functions
Definitions: f (x) = loga x (a > 0)
Derivative:
f ′(x) = (loga x)′ =
1
x ln a.
How to compute: Strictly speaking you should be using the law forinverse functions. But if you know already that (ln x)′ = 1/x , thenyou can simply make use of the fact loga x = ln x/ ln a.
Antiderivative:∫f (x)dx =
∫loga xdx =
1
ln a(x ln x − x) + C
How to compute: Use integration by parts to solve the special casethat a = e, then again use loga x = ln x/ ln a.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 5 / 16
![Page 14: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/14.jpg)
. . . . . .
Logarithm Functions
Definitions: f (x) = loga x (a > 0)
Derivative:
f ′(x) = (loga x)′ =
1
x ln a.
How to compute: Strictly speaking you should be using the law forinverse functions. But if you know already that (ln x)′ = 1/x , thenyou can simply make use of the fact loga x = ln x/ ln a.
Antiderivative:∫f (x)dx =
∫loga xdx =
1
ln a(x ln x − x) + C
How to compute: Use integration by parts to solve the special casethat a = e, then again use loga x = ln x/ ln a.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 5 / 16
![Page 15: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/15.jpg)
. . . . . .
Logarithm Functions
Definitions: f (x) = loga x (a > 0)
Derivative:
f ′(x) = (loga x)′ =
1
x ln a.
How to compute: Strictly speaking you should be using the law forinverse functions. But if you know already that (ln x)′ = 1/x , thenyou can simply make use of the fact loga x = ln x/ ln a.
Antiderivative:∫f (x)dx =
∫loga xdx =
1
ln a(x ln x − x) + C
How to compute: Use integration by parts to solve the special casethat a = e, then again use loga x = ln x/ ln a.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 5 / 16
![Page 16: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/16.jpg)
. . . . . .
Logarithm Functions
Definitions: f (x) = loga x (a > 0)
Derivative:
f ′(x) = (loga x)′ =
1
x ln a.
How to compute: Strictly speaking you should be using the law forinverse functions. But if you know already that (ln x)′ = 1/x , thenyou can simply make use of the fact loga x = ln x/ ln a.
Antiderivative:∫f (x)dx =
∫loga xdx =
1
ln a(x ln x − x) + C
How to compute: Use integration by parts to solve the special casethat a = e, then again use loga x = ln x/ ln a.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 5 / 16
![Page 17: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/17.jpg)
. . . . . .
Trigonometric functions
Definitions: sin x , cos x , tan x , cot x , sec x , csc x
Derivative:
(sin x)′ = cos x , (cos x)′ = − sin x ,
(tan x)′ = sec2 x , (cot x)′ = − csc2 x ,
(sec x)′ = sec x tan x , (csc x)′ = − csc x cot x .
How to compute: Use definitions of derivatives and the trigonometricidentities to work on sin x and cos x . Use laws of quotients to work ontan x and cot x . Use either law of quotients or chain rule to work onsec x and csc x .
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 6 / 16
![Page 18: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/18.jpg)
. . . . . .
Trigonometric functions
Definitions: sin x , cos x , tan x , cot x , sec x , csc x
Derivative:
(sin x)′ = cos x , (cos x)′ = − sin x ,
(tan x)′ = sec2 x , (cot x)′ = − csc2 x ,
(sec x)′ = sec x tan x , (csc x)′ = − csc x cot x .
How to compute: Use definitions of derivatives and the trigonometricidentities to work on sin x and cos x . Use laws of quotients to work ontan x and cot x . Use either law of quotients or chain rule to work onsec x and csc x .
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 6 / 16
![Page 19: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/19.jpg)
. . . . . .
Trigonometric functions
Definitions: sin x , cos x , tan x , cot x , sec x , csc x
Derivative:
(sin x)′ = cos x , (cos x)′ = − sin x ,
(tan x)′ = sec2 x , (cot x)′ = − csc2 x ,
(sec x)′ = sec x tan x , (csc x)′ = − csc x cot x .
How to compute: Use definitions of derivatives and the trigonometricidentities to work on sin x and cos x . Use laws of quotients to work ontan x and cot x . Use either law of quotients or chain rule to work onsec x and csc x .
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 6 / 16
![Page 20: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/20.jpg)
. . . . . .
Trigonometric functions
Definitions: sin x , cos x , tan x , cot x , sec x , csc x
Derivative:
(sin x)′ = cos x , (cos x)′ = − sin x ,
(tan x)′ = sec2 x , (cot x)′ = − csc2 x ,
(sec x)′ = sec x tan x , (csc x)′ = − csc x cot x .
How to compute: Use definitions of derivatives and the trigonometricidentities to work on sin x and cos x . Use laws of quotients to work ontan x and cot x . Use either law of quotients or chain rule to work onsec x and csc x .
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 6 / 16
![Page 21: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/21.jpg)
. . . . . .
Trigonometric functions
Antiderivative:∫sin xdx = − cos x + C ,
∫cos xdx = sin x + C ,∫
tan xdx = − ln | cos x |+ C ,
∫cot xdx = ln | sin x |+ C ,∫
sec xdx = ln | sec x + tan x |+ C ,
∫csc x = − ln | csc x + cot x |.
How to compute: Use the derivatives above to see the first two.Write in quotients and use substitutions then you will see the secondtwo. Use trigonometric techniques to get the last two.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 7 / 16
![Page 22: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/22.jpg)
. . . . . .
Trigonometric functions
Antiderivative:∫sin xdx = − cos x + C ,
∫cos xdx = sin x + C ,∫
tan xdx = − ln | cos x |+ C ,
∫cot xdx = ln | sin x |+ C ,∫
sec xdx = ln | sec x + tan x |+ C ,
∫csc x = − ln | csc x + cot x |.
How to compute: Use the derivatives above to see the first two.Write in quotients and use substitutions then you will see the secondtwo. Use trigonometric techniques to get the last two.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 7 / 16
![Page 23: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/23.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx
=
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) =
1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 24: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/24.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) =
1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 25: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/25.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x
=
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) =
1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 26: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/26.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) =
1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 27: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/27.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |)
=1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 28: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/28.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) =
1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣
It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 29: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/29.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) =
1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣It would be fine to end here. This is a correct answer. It just take a fewmore steps to get what we are looking for
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 8 / 16
![Page 30: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/30.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) = 1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣
=1
2ln
∣∣∣∣(1 + sin x)(1 + sin x)
(1− sin x)(1 + sin x)
∣∣∣∣ =1
2ln
∣∣∣∣(1 + sin x)2
cos2 x
∣∣∣∣= ln
∣∣∣∣1 + sin x
cos x
∣∣∣∣ = ln |sec x + tan x |
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 9 / 16
![Page 31: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/31.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) = 1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣=
1
2ln
∣∣∣∣(1 + sin x)(1 + sin x)
(1− sin x)(1 + sin x)
∣∣∣∣
=1
2ln
∣∣∣∣(1 + sin x)2
cos2 x
∣∣∣∣= ln
∣∣∣∣1 + sin x
cos x
∣∣∣∣ = ln |sec x + tan x |
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 9 / 16
![Page 32: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/32.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) = 1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣=
1
2ln
∣∣∣∣(1 + sin x)(1 + sin x)
(1− sin x)(1 + sin x)
∣∣∣∣ =1
2ln
∣∣∣∣(1 + sin x)2
cos2 x
∣∣∣∣
= ln
∣∣∣∣1 + sin x
cos x
∣∣∣∣ = ln |sec x + tan x |
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 9 / 16
![Page 33: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/33.jpg)
. . . . . .
Details of∫sec xdx
∫sec xdx =
∫1
cos xdx =
∫cos x
cos2 xdx
=
∫1
1− sin2 xd sin x =
∫1
2
(1
1− sin x+
1
1 + sin x
)d sin x
=1
2(ln |1 + sin x | − ln |1− sin x |) = 1
2ln
∣∣∣∣1 + sin x
1− sin x
∣∣∣∣=
1
2ln
∣∣∣∣(1 + sin x)(1 + sin x)
(1− sin x)(1 + sin x)
∣∣∣∣ =1
2ln
∣∣∣∣(1 + sin x)2
cos2 x
∣∣∣∣= ln
∣∣∣∣1 + sin x
cos x
∣∣∣∣ = ln |sec x + tan x |
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 9 / 16
![Page 34: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/34.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy ⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 35: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/35.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy ⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 36: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/36.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.
Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy ⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 37: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/37.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy ⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 38: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/38.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy
⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 39: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/39.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy ⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 40: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/40.jpg)
. . . . . .
Inverse Trigonometric functions
Definitions: arcsin x , arccos x , arctan x , arccotx .
Derivative:
(arcsin x)′ =1√
1− x2, (arccos x)′ = − 1√
1− x2,
(arctan x)′ =1
1 + x2, (arccotx)′ = − 1
1 + x2.
How to compute: Use the techniques dealing with inverse functions.Example of computing arctan x :
y = arctan x ⇒ x = tan y
⇒ dx = sec2 ydy = (1 + tan2 y)dy = (1 + x2)dy ⇒ dy
dx=
1
1 + x2.
Antiderivative: Not interesting at least in 244. So forget it.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 10 / 16
![Page 41: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/41.jpg)
. . . . . .
Hyperbolic trigonometric functions
Definitions:
sinh x =ex − e−x
2, cosh x =
ex + e−x
2, tanh x =
sinh x
cosh x, coth x =
cosh x
sinh x.
Derivative:(sinh x)′ = cosh x , (cosh x)′ = sinh x .
The rest two are left as exercises for product rule.
How to compute: Straightforward.
Antiderivative:∫sinh xdx = − cosh x + C ,
∫cosh xdx = sinh x + C .
The rest two are left as exercises for technique of substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 11 / 16
![Page 42: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/42.jpg)
. . . . . .
Hyperbolic trigonometric functions
Definitions:
sinh x =ex − e−x
2, cosh x =
ex + e−x
2, tanh x =
sinh x
cosh x, coth x =
cosh x
sinh x.
Derivative:(sinh x)′ = cosh x , (cosh x)′ = sinh x .
The rest two are left as exercises for product rule.
How to compute: Straightforward.
Antiderivative:∫sinh xdx = − cosh x + C ,
∫cosh xdx = sinh x + C .
The rest two are left as exercises for technique of substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 11 / 16
![Page 43: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/43.jpg)
. . . . . .
Hyperbolic trigonometric functions
Definitions:
sinh x =ex − e−x
2, cosh x =
ex + e−x
2, tanh x =
sinh x
cosh x, coth x =
cosh x
sinh x.
Derivative:(sinh x)′ = cosh x , (cosh x)′ = sinh x .
The rest two are left as exercises for product rule.
How to compute: Straightforward.
Antiderivative:∫sinh xdx = − cosh x + C ,
∫cosh xdx = sinh x + C .
The rest two are left as exercises for technique of substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 11 / 16
![Page 44: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/44.jpg)
. . . . . .
Hyperbolic trigonometric functions
Definitions:
sinh x =ex − e−x
2, cosh x =
ex + e−x
2, tanh x =
sinh x
cosh x, coth x =
cosh x
sinh x.
Derivative:(sinh x)′ = cosh x , (cosh x)′ = sinh x .
The rest two are left as exercises for product rule.
How to compute: Straightforward.
Antiderivative:∫sinh xdx = − cosh x + C ,
∫cosh xdx = sinh x + C .
The rest two are left as exercises for technique of substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 11 / 16
![Page 45: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/45.jpg)
. . . . . .
Hyperbolic trigonometric functions
Definitions:
sinh x =ex − e−x
2, cosh x =
ex + e−x
2, tanh x =
sinh x
cosh x, coth x =
cosh x
sinh x.
Derivative:(sinh x)′ = cosh x , (cosh x)′ = sinh x .
The rest two are left as exercises for product rule.
How to compute: Straightforward.
Antiderivative:∫sinh xdx = − cosh x + C ,
∫cosh xdx = sinh x + C .
The rest two are left as exercises for technique of substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 11 / 16
![Page 46: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/46.jpg)
. . . . . .
Hyperbolic trigonometric functions
Definitions:
sinh x =ex − e−x
2, cosh x =
ex + e−x
2, tanh x =
sinh x
cosh x, coth x =
cosh x
sinh x.
Derivative:(sinh x)′ = cosh x , (cosh x)′ = sinh x .
The rest two are left as exercises for product rule.
How to compute: Straightforward.
Antiderivative:∫sinh xdx = − cosh x + C ,
∫cosh xdx = sinh x + C .
The rest two are left as exercises for technique of substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 11 / 16
![Page 47: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/47.jpg)
. . . . . .
More formulas
∫1
a2 + x2=
1
aarctan
x
a+ C
How to compute: Substitution by scalar.
∫1
x2 − a2=
1
2aln
∣∣∣∣a− x
a+ x
∣∣∣∣+ C
How to compute: Either by trigonometric substitution or by breakingrational functions.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 12 / 16
![Page 48: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/48.jpg)
. . . . . .
More formulas
∫1
a2 + x2=
1
aarctan
x
a+ C
How to compute: Substitution by scalar.
∫1
x2 − a2=
1
2aln
∣∣∣∣a− x
a+ x
∣∣∣∣+ C
How to compute: Either by trigonometric substitution or by breakingrational functions.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 12 / 16
![Page 49: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/49.jpg)
. . . . . .
More formulas
∫1
a2 + x2=
1
aarctan
x
a+ C
How to compute: Substitution by scalar.
∫1
x2 − a2=
1
2aln
∣∣∣∣a− x
a+ x
∣∣∣∣+ C
How to compute: Either by trigonometric substitution or by breakingrational functions.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 12 / 16
![Page 50: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/50.jpg)
. . . . . .
More formulas
∫1
a2 + x2=
1
aarctan
x
a+ C
How to compute: Substitution by scalar.
∫1
x2 − a2=
1
2aln
∣∣∣∣a− x
a+ x
∣∣∣∣+ C
How to compute: Either by trigonometric substitution or by breakingrational functions.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 12 / 16
![Page 51: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/51.jpg)
. . . . . .
More formulas
∫1√
a2 − x2= arcsin
x
a+ C
How to compute: Again substitution by scalar.
∫1√
x2 ± a2= ln |x +
√x2 ± a2|+ C
How to compute: Either by trigonometric substitution or byhyperbolic substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 13 / 16
![Page 52: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/52.jpg)
. . . . . .
More formulas
∫1√
a2 − x2= arcsin
x
a+ C
How to compute: Again substitution by scalar.
∫1√
x2 ± a2= ln |x +
√x2 ± a2|+ C
How to compute: Either by trigonometric substitution or byhyperbolic substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 13 / 16
![Page 53: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/53.jpg)
. . . . . .
More formulas
∫1√
a2 − x2= arcsin
x
a+ C
How to compute: Again substitution by scalar.
∫1√
x2 ± a2= ln |x +
√x2 ± a2|+ C
How to compute: Either by trigonometric substitution or byhyperbolic substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 13 / 16
![Page 54: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/54.jpg)
. . . . . .
More formulas
∫1√
a2 − x2= arcsin
x
a+ C
How to compute: Again substitution by scalar.
∫1√
x2 ± a2= ln |x +
√x2 ± a2|+ C
How to compute: Either by trigonometric substitution or byhyperbolic substitution.
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 13 / 16
![Page 55: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/55.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 56: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/56.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.
∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 57: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/57.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx
=
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 58: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/58.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)
=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 59: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/59.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt
=
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 60: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/60.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt
=
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 61: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/61.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t=
ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 62: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/62.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣
= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 63: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/63.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣
= ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 64: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/64.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣
= ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 65: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/65.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 66: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/66.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C
= ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 67: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/67.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 − a2
Approach by trigonometric substitution: Let x = a sec t.∫1√
x2 − a2dx =
∫1
a√sec2t − 1
d( a
cos t
)=
∫1
tan t· sin t
cos2tdt =
∫1
cos tdt =
∫d sin t
1− sin2t= ln
∣∣∣∣1 + sin t
1− sin t
∣∣∣∣= ln
∣∣∣∣∣(1 + sin t)2
cos2t
∣∣∣∣∣ = ln
∣∣∣∣1 + sin t
cos t
∣∣∣∣ = ln | sec t + tan t|
= ln
∣∣∣∣∣∣xa +x√
1− (a/x)2
a
∣∣∣∣∣∣+ C = ln |x +√
x2 − a2|+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 14 / 16
![Page 68: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/68.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.
∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 69: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/69.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx
=
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 70: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/70.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 71: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/71.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt
=
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 72: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/72.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 73: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/73.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2
⇒ 2x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 74: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/74.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1
⇒ e2t − 2x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 75: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/75.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 76: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/76.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)
∫1√
x2 + a2dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 77: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/77.jpg)
. . . . . .
More detail about the last integral
Example: ∫1√
x2 + a2
Approach by hyperbolic substitution: Let x = a sinh t.∫1√
x2 + a2dx =
∫1
a√
sinh2t + 1d(a sinh t)
=
∫1
a cosh ta cosh tdt =
∫dt = t + C
x = a sinh t =et − e−t
2⇒ 2
x
aet = e2t − 1 ⇒ e2t − 2
x
aet − 1 = 0
⇒ et =x +
√x2 + a2
a(The smaller root makes et negative)∫
1√x2 + a2
dx = t + C = ln
∣∣∣∣∣x ±√x2 + a2
a
∣∣∣∣∣+ C
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 15 / 16
![Page 78: sites.math.rutgers.edufq15/244S14/Memo1-1.pdf · . . . . . . Logarithm Functions De nitions: f(x) = loga x (a > 0) Derivative: f′(x) = (log a x) ′ = 1 x lna: How to compute: Strictly](https://reader034.vdocuments.us/reader034/viewer/2022043014/5fb1f791b7cf0a2a2b044396/html5/thumbnails/78.jpg)
. . . . . .
The End
Fei Qi (Rutgers University) Review of Formulas in Calculus January 21, 2014 16 / 16