# NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9

These NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 7 Integrals Exercise 7.9

Question 1.
$$\int_{-1}^{1}(x+1)$$
Solution:
Let I = $$\int_{-1}^{1}(x+1)$$ dx
= $$\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}$$
= $$\left(\frac{1}{2}+1\right)-\left(\frac{1}{2}-1\right)=\frac{3}{2}+\frac{1}{2}=2$$

Question 2.
$$\int _{ 2 }^{ 3 }{ \frac { 1 }{ x } dx }$$
Solution:
Let I = $$\int_{2}^{3} \frac{1}{x} d x=[\log |x|]_{2}^{3}$$
= $$\log 3-\log 2=\log \frac{3}{2}$$

Question 3.
$$\int _{ 1 }^{ 2 }{ \left( { 4x }^{ 3 }-{ 5x }^{ 2 }+6x+9 \right) dx }$$
Solution:

Question 4.
$$\int_{0}^{\frac{\pi}{4}} \sin 2 x$$
Solution:

Question 5.
$$\int_{0}^{\frac{\pi}{2}} \cos 2 x$$
Solution:

Question 6.
$$\int _{ 4 }^{ 5 }{ { e }^{ x }dx }$$
Solution:
Let I = $$\int _{ 4 }^{ 5 }{ { e }^{ x }dx }$$
= $$\left[e^{x}\right]_{4}^{5}=e^{5}-e^{4}$$
= e4 (e – 1)

Question 7.
$$\int_{0}^{\frac{\pi}{4}} \tan x$$
Solution:
$$\int_{0}^{\frac{\pi}{4}} \tan x d x$$
= $$\left[\log |\sec x|_{0}^{\frac{x}{4}}\right.$$
= $$\log \left|\sec \frac{\pi}{4}\right|-\log |\sec 0|=\log \sqrt{2}-\log 1=\log \sqrt{2}$$

Question 8.
$$\int_{\frac{π}{6}}^{\frac{π}{4}}$$ cosec x dx
Solution:

Question 9.
$$\int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } }$$
Solution:
Let I = $$\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}=\left[\sin ^{-1} x\right]_{0}^{1}$$
= $$\sin ^{-1}(1)-\sin ^{-1}(0)=\frac{\pi}{2}-0=\frac{\pi}{2}$$

Question 10.
$$\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } }$$
Solution:

Question 11.
$$\int _{ 2 }^{ 3 }{ \frac { dx }{ { x }^{ 2 }-1 } }$$
Solution:

Question 12.
$$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 } } x$$
Solution:

Question 13.
$$\int _{ 2 }^{ 3 }{ \frac { x }{ { x }^{ 2 }+1 } }$$
Solution:

Question 14.
$$\int _{ 0 }^{ 1 }{ \frac { 2x+3 }{ { 5x }^{ 2 }+1 } }$$
Solution:

Question 15.
$$\int _{ 0 }^{ 1 }{ { xe }^{ { x }^{ 2 } }dx }$$
Solution:

Question 16.
$$\int _{ 1 }^{ 2 }{ \frac { { 5x }^{ 2 } }{ { x }^{ 2 }+4x+3 } dx }$$
Solution:

20x + 15A(x + 3) + B(x + 1) …. (2)
Put x = – 1 in (2), we get
– 5 = 2A ∴ A = $$\frac { – 5 }{ 2 }$$
Put x = – 3 in (2), we get
– 45 = – 2B ∴ B = $$\frac { 45 }{ 2 }$$

Question 17.
$$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( { 2sec }^{ 2 }x+{ x }^{ 3 }+2 \right) dx }$$
Solution:

Question 18.
$$\int _{ 0 }^{ \pi }{ \left( { sin }^{ 2 }\frac { x }{ 2 } -{ cos }^{ 2 }\frac { x }{ 2 } \right) }$$dx
Solution:

Question 19.
$$\int _{ 0 }^{ 2 }{ \frac { 6x+3 }{ { x }^{ 2 }+4 } } dx$$
Solution:

Question 20.
$$\int _{ 0 }^{ 1 }{ \left( { xe }^{ x }+sin\frac { \pi x }{ 4 } \right) dx }$$
Solution:

Question 21.
$$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$$ equals
(a) $$\frac { \pi }{ 3 }$$
(b) $$\frac { 2\pi }{ 3 }$$
(c) $$\frac { \pi }{ 6 }$$
(d) $$\frac { \pi }{ 12 }$$
Solution:
(d) $$\frac { \pi }{ 12 }$$
Let I = $$\int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$$
= $$\left[\tan ^{-1} x\right]_{1}^{\sqrt{3}}=\tan ^{-1} \sqrt{3}-\tan ^{-1} 1$$
= $$\frac{\pi}{3}-\frac{\pi}{4}=\frac{\pi}{12}$$

Question 22.
$$\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$$ equals
(a) $$\frac { \pi }{ 6 }$$
(b) $$\frac { \pi }{ 12 }$$
(c) $$\frac { \pi }{ 24 }$$
(d) $$\frac { \pi }{ 4 }$$
Solution:
(c) $$\frac { \pi }{ 24 }$$

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