more u-substitution

More U-Substitution

Chapter 5.5

February 8, 2007

In-Class Assignment

Integrate using two different methods:

1st by multiplying out and integrating 2nd by u-substitution

Do you get the same result? (Don’t just assume or claim you do; multiply out your results to show it!)

If you don’t get exactly the same answer, is it a problem? Why or why not?

3x −1( )∫2dx

Fundamental Theorem of Calculus (Part 2)

If f is continuous on [a, b], then :

Where F is any antiderivative of f. ( )

f (t)dta

b

∫ =F(b)−F(a)

F ' = f

Substitution Rule for Indefinite Integrals

If u = g(x) is a differentiable function whose range is an interval I and f is continuous on I, then

f ' g(x)( ) g'(x)dx=∫ f u( )∫ du= f g x( )( ) +C

Substitution Rule for Indefinite Integrals

If g’(x) is continuous on [a,b] and f is continuous on the range of u = g(x), then

f g(x)( ) g'(x)dxa

b

∫ = f(u)dug(a)

g(b)

∫

Using the Chain Rule, we know that:

d

dxsin x2( )( ) =2xcos x2( )

Evaluate:

Evaluate:

Looks almost like cos(x2) 2x, which is

the derivative of sin(x2).


d


Evaluate:

We will rearrange the integral to get an exact match:

d



Evaluate:



d


x cos x2( )dx= cos x2( )xdx∫∫

Evaluate:



d


x cos x2( )dx= cos x2( )xdx∫∫=12

cos x2( )2xdx∫

Evaluate:


We put in a 2 so the pattern will match.


d



cos x2( )2xdx∫

Evaluate:


We put in a 2 so the pattern will match.

So we must also put in a 1/2 to keep the

problem the same.


d



cos x2( )2xdx∫

Evaluate:



d



cos x2( )2xdx∫

Evaluate:



d



cos x2( )2xdx∫

=12sin x2( )⎡⎣ ⎤⎦+C

Check Answer:

Check:

Check Answer:

Check:

From the chain rule

Check Answer:

Indefinite Integrals by Substitution

1) Choose u.


1) Choose u.

2) Calculate du.

€

du =du

dxdx


1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)€

du =du

dxdx


1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)

4) Solve the new integral.

€

du =du

dxdx



1) Choose u.

2) Calculate du.

3) Substitute u.Arrange to have du in your integral also.(All xs and dxs must be replaced!)

4) Solve the new integral.

5) Substitute back in to get x again.

€

du =du

dxdx

ExampleA linear substitution:

Let u = 3x + 2. Then du = 3dx.

e3x+2dx∫ =13

e3x+23dx∫

=1

3eudu∫

=1

3eu + C

=1

3e3x+2 + C

Choosing u Try to choose u to be an inside function.

(Think chain rule.) Try to choose u so that du is in the

problem, except for a constant multiple.

Choosing uFor

u = 3x + 2 was a good choice because

(1) 3x + 2 is inside the exponential.

(2) The derivative is 3, which is only a constant.

PracticeLet u =

du =

PracticeLet u = x2 + 1

du =


du = 2x dx


du = 2x dx

Make this a 2x dx and we’re

set!


du = 2x dx


du = 2x dx

= 1

2ln x2 + 1 + C

PracticeLet u =

du =

PracticeLet u = sin(x)

du =

PracticeLet u = sin(x)

du = cos(x) dx

PracticeLet u =

du = An alternate possibility:

PracticeLet u = cos(x)

du = –sin(x) dxAn alternate possibility:

Practice

Note:

Practice

Note:

What’s the difference?

PracticeNote:


Practice

Note:


Practice

Note:


This is 1!

Practice

Note:


Practice

Note:


That is, the difference is a constant.

PracticeLet u =

du =

PracticeLet u = 1+ex

du =

PracticeLet u = 1+ex

du = exdx

PracticeLet u =

du =

PracticeLet u = x2

du = 2x dx

Doesn’t Fit All

Doesn’t Fit AllWe can’t use u–substitution to solveeverything.

Doesn’t Fit AllWe can’t use u–substitution to solveeverything. For example:

Let u = du =


Let u = x2

du = 2x dx


Let u = x2

du = 2x dx

We need 2x this time, not just 2.


Let u = x2

du = 2x dx


We CANNOT multiply by a variable to adjust our integral.


Let u = x2

du = 2x dx


We CANNOT multiply by a variable to adjust our integral.

We cannot complete this problem

Doesn’t Fit AllFor the same reason, we can’t do the following by u–substitution:

Doesn’t Fit AllFor the same reason, we can’t do the following by u–substitution:

But we already knew how to do this!

Morals No one technique works for everything. Don’t forget things we already know!

There are lots of integrals we will never learn how to solve…

That’s when Simpson’s Rule, the Trapezoidal Rule, Midpoint,….. need to be used to estimate…...

Use u-substitution to find:

Similarly for:

tan xdx∫sin x

cos xdx∫ ln sec x +C

cot xdx∫cos x

sin xdx∫ ln sin x +C

Definite Integral Examples:

e1

x

x2dx

1

2

∫

xe−x2dx0

1

∫

sin−1(x)

1−x2dx

0

12

∫

e1

x

x2dx

1

2

∫

xe−x2dx0

1

∫

sin−1(x)

1−x2dx

0

12

∫

e− e

1

21−

1e

⎛⎝⎜

⎞⎠⎟

π 2

72

eudu1

2

1

∫

1

2eudu

−1

0

∫

udu0

π6

∫

Definite Integrals

Evaluate: dx

x −2( )20

4

∫

Definite Integrals: Why change bounds?

Can simplify calculations. (no need to substitute back to the original variable)

Try:

Change of bounds: when x = -5, u = 25 - (-5)2=0 when x = 5, u = 25 - (5)2 = 0

x 25 −x2dx−5

5

∫ −1

225 − x2 (−2x)dx

−5

5

∫u =25 −x2

du=−2xdx

−1

2udu

0

0

∫

Definite Integrals & bounds

x 25 −x2dx−5

5

∫ −1

225 − x2 (−2x)dx

−5

5

∫

u =25 −x2

du=−2xdx

−1

2udu

0

0

∫

Examples to try:

cos(3x)dx∫

x 4 + x2( )dx∫

sin( x )

xdx∫

ln x( )( )2

xdx∫

Examples

cos(3x)dx∫ 1

3cos( 3x ) 3dx∫

x 4 + x2( )dx∫

sin( x )

xdx∫

ln x( )( )2

xdx∫

1

24 + x2( ) 2xdx∫

2sin( x )

2 xdx∫

ln(x)( )2

xdx∫

1

3cos( u ) du∫

1

2u du∫

2 sin( u ) du∫

u( )2du∫

Summary Make a u-substitution: find u and du, then transform

to an integral we can do. Be sure to transform back to the original variable!

Rearrange to get du Substitution will still not solve every integral. (Nor is it

always needed!)

Change of Variable

Applet

more u-substitution

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