exercise 13c page: 678 - byju's...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] taking first...
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![Page 1: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/1.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
Exercise 13C page: 678 Evaluate the following integrals:
1. ∫ x ex dx
Solution:
It is given that
∫ x ex dx
We can write it as
By integrating w.r.t x
= x ex - ∫ 1 ex dx
So we get
= x ex – ex + c
By taking ex as common
= ex (x – 1) + c
2. ∫ x cos x dx
Solution:
It is given that
∫ x cos x dx
We can write it as
By integrating w.r.t x
= x sin x - ∫ 1 sin x dx
So we get
= x sin x + cos x + c
3. ∫ x e2x dx
Solution:
It is given that
∫ x e2x dx
We can write it as
![Page 2: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/2.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
4. ∫ x sin 3x dx
Solution:
It is given that
∫ x sin 3x dx
We can write it as
5. ∫ x cos 2x dx
Solution:
It is given that
∫ x cos 2x dx
We can write it as
So we get
![Page 3: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/3.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
6. ∫ x log 2x dx
Solution:
It is given that
∫ x log 2x dx
We can write it as
7. ∫ x cosec2 x dx
Solution:
It is given that
∫ x cosec2 x dx
We can write it as
By integrating w.r.t x
= x (- cot x) - ∫ 1 (- cot x) dx
So we get
![Page 4: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/4.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
= - x cot x + ∫ cot x dx
Again by integration we get
= - x cot x + log |sin x| + c
8. ∫ x2 cos x dx
Solution:
It is given that
∫ x2 cos x dx
We can write it as
By integrating w.r.t x
= x2 sin x - ∫ 2x sin x dx
So we get
= x2 sin x – 2 [∫ x sin x dx]
Now apply by the part method
= x2 sin x – 2 ∫ x sin x dx
Again by integration
We get
= x2 sin x – 2 [x (- cos x) - ∫ 1. (- cos x) dx]
By integrating w.r.t x
= x2 sin x – 2 (- x cos x + sin x) + c
On further simplification
= x2 sin x + 2x cos x – 2 sin x + c
9. ∫ x sin2 x dx
Solution:
It is given that
∫ x sin2 x dx
We can write it as
sin2 x = (1 + cos 2x)/ 2
By integration
It can be written as
![Page 5: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/5.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
10. ∫ x tan2 x dx
Solution:
It is given that
∫ x tan2 x dx
We can write it as
tan2 x = sec2 x – 1
So we get
∫ x tan2 x dx = ∫ x ( sec2 x – 1) dx
By further simplification
= ∫ x sec2 x dx - ∫ x dx
Taking first function as x and second function as sec2 x
∫ x sec2 x dx - ∫ x dx
We get
![Page 6: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/6.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
By integration w.r.t x
= {x tan x - ∫ tan x dx} – x2/ 2
It can be written as
= x tan x – log |sec x| - x2/ 2 + c
So we get
= x tan x – log |1/cos x| - x2/ 2 + c
Here
= x tan x + log |cos x| - x2/ 2 + c
11. ∫ x2 ex dx
Solution:
It is given that
∫ x2 ex dx
By integrating w.r.t x
We get
= x2 ex – 2 [x ex - ∫1. ex dx]
By further simplification
= x2 ex – 2 [x ex – ex] + c
By multiplication
= x2 ex – 2x ex + 2 ex + c
By taking ex as common
= ex (x2 – 2x + 2) + c
13. ∫ x2 e3x dx
Solution:
It is given that
∫ x2 e3x dx
We can write it as
![Page 7: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/7.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
14. ∫ x2 sin2 x dx
Solution:
It is given that
∫ x2 sin2 x dx
We know that
![Page 8: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/8.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
Now by integrating the second part we get
![Page 9: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/9.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
15. ∫ x3 log 2x dx
Solution:
It is given that
∫ x3 log 2x dx
We can write it as
16. ∫ x log (x + 1) dx
Solution:
It is given that
![Page 10: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/10.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
Solution:
It is given that
![Page 11: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/11.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
Solution:
It is given that
We can write it as
= t et - ∫1. et dt
So we get
= t et – et + C
Now by replacing t with x2
![Page 12: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/12.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
19. ∫ x sin3 x dx
Solution:
It is given that
∫ x sin3 x dx
We know that
sin 3x = 3 sin x – 4 sin3 x
It can be written as
sin3 x = (3 sin x – sin 3x)/4
We get
So we get
![Page 13: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/13.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
20. ∫ x cos3 x dx
Solution:
It is given that
∫ x cos3 x dx
We know that
cos3 x = (cos 3x + 3 cos x)/4
It can be written as
21. ∫ x3 cos x2 dx
Solution:
It is given that
∫ x3 cos x2 dx
We can write it as
∫ x x2 cos x2 dx
Take x2 = t
So we get 2x dx = dt
x dx = dt/2
![Page 14: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/14.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
22. ∫ sin x log (cos x) dx
Solution:
It is given that
∫ sin x log (cos x) dx
Consider first function as log (cos x) and second function as sin x
So we get
= - cos x log (cos x) - ∫ sin x dx
Again by integrating the second term
= - cos x log (cos x) + cos x + c
23. ∫ x sin x cos x dx
Solution:
It is given that
∫ x sin x cos x dx
We know that
sin 2x = 2 sin x cos x
It can be written as
Consider first function as x and second function as sin 2x
![Page 15: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/15.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
24. ∫ cos √x dx
Solution:
It is given that
∫ cos √x dx
Take √x = t
So we get
1/2√x dx = dt
By cross multiplication
dx = 2 √x dt
Here dx = 2t dt
It can be written as
∫ cos √x dx = 2 ∫t cos t dt
Take first function as t and second function as cos t
By integrating w.r.t t
= 2 (t sin t - ∫ sin t dt)
We get
= 2t sin t + 2 cos t + c
Now by substituting t as √x
= 2√x sin √x + 2 cos √x + c
Taking 2 as common
= 2 (cos √x + √x sin √x) + c
25. ∫ cosec3 x dx
Solution:
It is given that
∫ cosec3 x dx
We can write it as
![Page 16: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/16.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
∫ cosec3 x dx = ∫ cosec x cosec2 x dx
So we get
By integration we get
= cosec x (- cot x) - ∫ (- cosec x cot x) (- cot x) dx
It can be written as
= - cosec x cot x - ∫ cosec x cot2 x dx
Here cot2 x = cosec2 x – 1
We get
= - cosec x cot x - ∫ cosec x (cosec2 x – 1) dx
By further simplification
= - cosec x cot x - ∫ cosec3 x dx + ∫ cosec x dx
We know that ∫ cosec3 x dx = 1
∫cosec3 x dx = - cosec x cot x - ∫ cosec3 x dx + ∫ cosec x dx
On further calculation
2 ∫cosec3 x dx = - cosec x cot x + ∫ cosec x dx
By integration we get
2 ∫cosec3 x dx = - cosec x cot x + log |tan x/2| + c
Here
∫cosec3 x dx = - ½ cosec x cot x + ½ log |tan x/2| + c
27. ∫ sin x log (cos x) dx
Solution:
It is given that
∫ sin x log (cos x) dx
Take cos x = t
So we get
- sin x dx = dt
It can be written as
∫ sin x log (cos x) dx = - ∫log t dt = - ∫1. log t dt
Consider first function as log t and second function as 1
By integrating w.r.t t
= - log t. t + ∫1/t dt
Again by integrating the second term
= - t log t + t + c
Now replace t as cos x
= t (- log t + 1) + c
We get
= cos x (1 – log (cos x)) + c
![Page 17: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/17.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
Solution:
Take log x = t
So we get
1/x dx = dt
Here
Again by integrating the second term we get
= t log t – t + c
Replace t with log x
= log x. log (log x) – log x + c
By taking log x as common
= log x (log (log x) – 1) + c
29. ∫ log (2 + x2) dx
Solution:
It is given that ∫ log (2 + x2) dx
We can write it as
= ∫1. log (2 + x2) dx
Consider first function as log (2 + x2) and second function as 1
![Page 18: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/18.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
So we get
Solution:
Consider log x = t
Where x = et
So we get
dx = et dt
We can write it as
![Page 19: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/19.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
32. ∫ e –x cos 2x cos 4x dx
Solution:
Using the formula
cos A cos B = ½ [cos (A + B) + cos (A – B)]
We get
cos 4x cos 2x = ½ [cos (4x + 2x) + cos (4x – 2x)] = ½ [cos 6x + cos 2x]
By applying in the original equation
∫ e –x cos 2x cos 4x dx = ∫ e –x (½ [cos 6x + cos 2x])
= ½ [∫e –x cos 6x dx + ∫e –x cos 2x dx]
Taking first function as cos 6x and cos 2x and second function as e –x
Now by solving both parts separately
![Page 20: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/20.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
![Page 21: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/21.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
33. ∫e √x dx
Solution:
Consider √x = t
We get
It can be written as
![Page 22: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/22.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
dx = 2 √x dt
where dx = 2t dt
We can replace it in the equation
∫e √x dx = ∫et 2t dt
So we get
= 2 ∫t et dt
Consider the first function as t and second function as et
By integration w.r.t. t
= 2 (t et - ∫1. et dt)
We get
= 2 (t et – et) + c
Taking et as common
= 2 et (t – 1) + c
Substituting the value of t we get
= 2 e √x (√x – 1) + c
34. ∫ e sin x sin 2x dx
Solution:
We know that
sin 2x = 2 sin x cos x
So we get
∫ e sin x sin 2x dx = 2 ∫ e sin x sin x cos x dx
Take sin x = t
So we get cos x dx = dt
It can be written as
2 ∫ e sin x sin x cos x dx = 2 ∫ e t t dt
Consider first function as t and second function as et
By integrating w.r.t. t
= 2 (t et - ∫1. et dt)
We get
= 2 (t et – et) + c
Here
= 2 et (t – 1) + c
By substituting the value of t
= 2 esin x (sin x – 1) + C
![Page 23: Exercise 13C page: 678 - BYJU'S...= ½ [∫e – x cos 6x dx + ∫e – cos 2x dx] Taking first function as cos 6x and cos 2x and second function as e –x Now by solving both parts](https://reader033.vdocuments.us/reader033/viewer/2022060813/6091af253889841ce13a8ff4/html5/thumbnails/23.jpg)
RS Aggarwal Solutions for Class 12 Maths Chapter 13 –
Methods of Integration
Solution:
Take sin -1 x = t
Here x = sin t
So we get
By integration we get
= t (- cos t) - ∫1. (- cos t) dt
We get
= - t cos t + sin t + c
It can be written as
cos t = √1 – sin2 t
Here we get
= - t (√1 – sin2 t) + sin t + c
By replacing t as sin-1 x
= - sin-1 x (√1 – x2) + x + c