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) \(\frac { \pi }{ 24 }\)