Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.3
![Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.3 Class 12 Maths NCERT Solutions for Chapter 7 Integrals Exercise 7.3](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiz8dk79RmrWE-DY8thaj39zXxSvP_iyWl-dbpT2QGxW7ASrWIPy-EiLONMqTm8r0GVpjaghmq0ms_WJdFvGHN0LDFVvzObtWg8g-Y8tA6w6i6DtGr4749II3XVa6Ep_xCqXXJcMt6pxoP3dTTUHaqCAkzFhVQ4_XOYN6U0odXWMFn88ZbUUNKuu6sh/w647-h305-rw/cbse-class-12-maths-ncert-solutions-chapter-7-exercise-7-3.jpg)
Integrals Exercise 7.3 Solutions
1. Find the integrals of the functions sin2 (2x + 5)
Solution
2. Find the integrals of the functions sin 3x cos 4x.
Solution
It is known that, sin A cos B = (1/2) {sin (A + B) + sin (A - B)}
3. Find the integrals of the functions cos 2x cos 4x cos 6x .
Solution
It is known that, cos A cos B = (1/2) {cos (A + B) + cos (A - B)}
4. Find the integrals of the functions sin3 (2x + 1).
Solution
Let I = ∫ sin3 (2x + 1)
⇒ ∫sin3 (2x + 1) dx = ∫ sin2 (2x + 1) . sin(2x + 1) dx
= ∫[1 - cos2 (2x + 1)] sin(2x + 1)dx
⇒ ∫sin3 (2x + 1) dx = ∫ sin2 (2x + 1) . sin(2x + 1) dx
= ∫[1 - cos2 (2x + 1)] sin(2x + 1)dx
Let cos(2x + 1) = t
⇒ -2sin(2x + 1)dx = dt
⇒ sin(2x + 1)dx = -dt/2
⇒ -2sin(2x + 1)dx = dt
⇒ sin(2x + 1)dx = -dt/2
5. Find the integrals of the functions sin3x cos3x
Solution
Let I = ∫ sin3 x cos3 x dx
= ∫ cos3 x . sin2 x . sin x . dx
= ∫ cos3 x(1 - cos2 x) sin x . dx
Let cos x = t
⇒ - sin x. dx = dt
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgzOJUdfaq14M2Zm0nmY-Ea7WXRreIV62Uopc0b8yvy3Bgy4Bs9xbASDzVlmqCyYIiK-eurIS5LiVJrqdaX2jQUCIk0AXggyUiW99r9lwfVcC-d_UFlgv4ZCZCfr67OQxOkjlZrv_EkqR0G-9lOXAq2yYKKug16D_HL_nU8CCKt_ElaO1dTw8OA-iWq/w204-h225-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.3%20img%205.JPG)
= ∫ cos3 x . sin2 x . sin x . dx
= ∫ cos3 x(1 - cos2 x) sin x . dx
Let cos x = t
⇒ - sin x. dx = dt
6. Find the integrals of the functions sin x sin 2x sin 3x
Solution
It is known that, sin A sin B = (1/2) {cos (A - B) - cos (A + B)} ![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiGpX6fZ4NUblGwbeucbT1SxtqwG1Kvi8_4eOTTGpe5qPtEbhNXh0Do6yJlB8rT6vElRij_2KRp-Q_HY-CSQEc5HWKmh0xQNniqn0dpDGGeUqfdBAwZfCRXNFSMpkRhTx-xMa52gCUrtLKeAShcSpT6a9eBeSC3mKnbwPe_uTMgRBgtKz1ph7If9UuU/w459-h402-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.3%20img%206.JPG)
7. Find the integrals of the functions sin 4x sin 8x.
Solution
It is known that, sin A sin B = (1/2) {cos (A - B) - cos (A + B)}
8. Find the integrals of the functions (1 - cos x)/(1 + cos x)
Solution
9. Find the integrals of the functions cos x/(1 + cos x).
Solution
10. Find the integrals of the functions sin4 x
Solution
sin4 x = sin2 x sin2 x
11. Find the integrals of the functions cos4 2x
Solution
12. Find the integrals of the functions sin2 x/(1 + cos x)
Solution
13. Find the integrals of the functions (cos 2x - cos 2α)/(cos x - cos α)
Solution
14. Find the integrals of the functions (cos x - sin x)/(1 + sin 2x)
Solution
15. Find the integrals of the functions tan3 2x sec 2x.
Solution
tan32x sec 2x = tan2 2x tan 2x sec 2x
= (sec2 2x - 1) tan 2x sec 2x
= sec2 2x. tan 2x sec 2x - tan 2x sec2x
∴ ∫tan3 2x sec 2x dx = ∫ sec2 2x tan 2x sec 2x dx - ∫ tan 2x sec 2x dx
= ∫ sec2 2x tan 2x sec 2x dx - sec 2x/2 + C
Let sec 2x = t
∴ 2 sec 2x tan 2x dx = dt
= (sec2 2x - 1) tan 2x sec 2x
= sec2 2x. tan 2x sec 2x - tan 2x sec2x
∴ ∫tan3 2x sec 2x dx = ∫ sec2 2x tan 2x sec 2x dx - ∫ tan 2x sec 2x dx
= ∫ sec2 2x tan 2x sec 2x dx - sec 2x/2 + C
Let sec 2x = t
∴ 2 sec 2x tan 2x dx = dt
16. Find the integrals of the functions tan4x .
Solution
tan4 x
= tan2 x . tan2 x
= (sec2 x - 1) tan2 x
= sec2 x tan2 x - tan2 x
= sec2 x tan2 x - (sec2 x - 1)
= sec2 x tan2 x - sec2 x + 1
= tan2 x . tan2 x
= (sec2 x - 1) tan2 x
= sec2 x tan2 x - tan2 x
= sec2 x tan2 x - (sec2 x - 1)
= sec2 x tan2 x - sec2 x + 1
∴ ∫ tan4 x dx = ∫ sec2 x tan2 x dx - ∫ sec2 x dx + ∫ 1 . dx
= ∫ sec2 x tan2 x dx - tan x + x + C ...(1)
Consider ∫ sec2 x tan2 x dx
Let tan x = t ⇒ sec2 x dx = dt
⇒ ∫ sec2 x tan2 x dx = ∫t2 dt = t3/3 = tan3 x/3
= ∫ sec2 x tan2 x dx - tan x + x + C ...(1)
Consider ∫ sec2 x tan2 x dx
Let tan x = t ⇒ sec2 x dx = dt
⇒ ∫ sec2 x tan2 x dx = ∫t2 dt = t3/3 = tan3 x/3
From equation (1), we obtain
∫ tan4 x dx = (1/3) tan3 x - tan x + x + C
∫ tan4 x dx = (1/3) tan3 x - tan x + x + C
17. Find the integrals of the functions (sin3 x + cos3 x)/(sin2 x cos2 x) .
Solution
18. Find the integrals of the functions (cos 2x + 2 sin2 x)/(cos2 x)
Solution
19. Find the integrals of the functions 1/(sin x cos3 x) .
Solution
Solution
21. Find the integrals of the functions sin-1 (cos x).
Solution
sin-1 (cos x)
Let cos x = t
Then, sin x = √(1 - t2 )
⇒ (- sin x)dx = dt
![](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdxKC0Ln3ZXjGh66jWSin21lbhe2NQZ1CmYKA_Uy_6GlL2OJ1ibfdHiIHsNQCACkxhEHrs6xDdVrm1EGWMa7chFMoWRilLj343km5x_UMMy-ipXU1gpTBN9dstek9BXUyH2vkxDX5tOdUB0UGTRWYybQDRhFqlljYpuiixpirqbJGg2FYy4L93Cm0a/w278-h864-rw/NCERT%20Solutions%20for%20Chapter%207%20Integrals%20Class%2012%20Maths%20Exercise%20-%207.3%20img%2020.JPG)
It is known that,
sin-1 x + cos-1 x = π/2
Let cos x = t
Then, sin x = √(1 - t2 )
⇒ (- sin x)dx = dt
It is known that,
sin-1 x + cos-1 x = π/2
22. Find the integrals of the functions 1/[cos(x - a) cos(x - b)] .
Solution
23. Choose the correct answer ∫(sin2 x - cos2 x)/(sin2 x cos2 x) dx
A. tan x + cot x + C
B. tan x + cosec x + C
C. −tan x + cot x + C
D. tan x + sec x + C
A. tan x + cot x + C
B. tan x + cosec x + C
C. −tan x + cot x + C
D. tan x + sec x + C
Solution
= tan x + cot x + C
Hence, the correct answer is A.
Hence, the correct answer is A.
24. Choose the correct answer ∫[ex (1 + x)]/[cos2 (ex x)] dx
(A) -cot (ex x) + C
(B) tan (xex ) + C
(C) tan (ex ) + C
(D) cot(ex ) + C
(A) -cot (ex x) + C
(B) tan (xex ) + C
(C) tan (ex ) + C
(D) cot(ex ) + C
Solution
Let ex. x = t
⇒ (ex. x + ex.1) dx = dt
ex (x + 1)dx = dt
Hence, the correct answer is B.