NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

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

NCERT Solutions for Class 12 Maths Chapter 7 Integrals Miscellaneous Exercise

Question 1.
\(\frac{1}{x-x^{3}}\)
Solution:
\(\int \frac{1}{x-x^{3}} d x=\int \frac{1}{x\left(1-x^{2}\right)} d x=\int \frac{1}{x(1-x)(1+x)} d x\)
Let\(\frac{1}{x(1-x)(1+x)}=\frac{\mathrm{A}}{x}+\frac{\mathrm{B}}{1-x}+\frac{\mathrm{C}}{1+x}\)
1 = A(1 – x)(l + x) + Bx(l + x) + Cx(l – x) Put x = 0 in (1), we get A = 1
Put x = 1 in (1), we get 1 = 2B ∴ B = \(\frac { 1 }{ 2 }\)
Put x = – 1 in (1), we get 1 = – 2C ∴ C = \(\frac { – 1 }{ 2 }\)

Question 2.
\(\frac{1}{\sqrt{x+a}+\sqrt{x+b}}\)
Solution:

Question 3.
\(\frac{1}{x \sqrt{a x-x^{2}}}\)
Solution:

Question 4.
\(\frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}\)
Solution:

Question 5.
\(\frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}\)
Solution:

Question 6.
\(\frac{5 x}{(x+1)\left(x^{2}+9\right)}\)
Solution:
Let \(\frac{5 x}{(x+1)\left(x^{2}+9\right)}\) = \(\frac{A}{x+1}+\frac{B x+C}{x^{2}+9}\)
⇒ 5x = A(x² + 9) + (Bx + C)(x + 1) … (1)
Put x = – 1 in (1), we get
– 5 = – 10
∴ A = \(\frac { -5 }{ 10 }\) = \(\frac { – 1 }{ 2 }\)
Equating the coefficients of x² and constant
term, we get A + B = 0, 9 A + C = 0

Question 7.
\(\frac{\sin x}{\sin (x-a)}\)
Solution:

= sin a ∫cot dt + cos a ∫dt
= sin a log |sin t| + cos a (t) + C₁
= sin a log |sin(x – a)| + cos a[x – 1] + C₁
= sin a.logsin (x – a) + x cosa + C, where C = C₁ – a cos a

Question 8.
\(\frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}\)
Solution:

Question 9.
\(\frac{\cos x}{\sqrt{4-\sin ^{2} x}}\)
Solution:

Question 10.
\(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\)
Solution:
Let I = \(\frac{\sin ^{8} x-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}\)dx
sin⁸x – cos⁸x (sin⁴x – cos⁴x)(sin⁴x + cos⁴x)
= (sin²x – cos²x)(sin²x + cos²x)(sin⁴x + cos⁴x)
= (sn²x – cos²x(1)[(sin²x + cos²x) – 2sin²x cos²x]
= (sin²x – cos²x)( 1 – 2sin²x cos²x)

Question 11.
\(\frac{1}{\cos (x+a) \cos (x+b)}\)
Solution:

Question 12.
\(\frac{x^{3}}{\sqrt{1-x^{8}}}\)
Solution:

Question 13.
\(\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}\)
Solution:
Let I = \(\frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}\) dx
Put t = e^x
\(\frac { dt }{ dx }\) = e^x
dt = e^x dx
= \(\int \frac{d t}{(1+t)(2+t)}\)
Let \(\frac{1}{(1+t)(2+t)}=\frac{\mathrm{A}}{(1+t)}+\frac{\mathrm{B}}{(2+t)}\)
∴ 1 = A(2 + t) + B(1 + t) … (1)
Put t = – 2 in (1), we get A = 1
Put t = – 2 in (1), we get 1 = – B ∴B = – 1
\(\frac{1}{(1+t)(2+t)}=\frac{1}{(1+t)}-\frac{1}{(2+t)}\)
I = \(\int \frac{1}{1+t} d t-\int \frac{1}{2+t} d t\)
= \(\log |1+t|-\log |2+t|+C\)
= \(\log \left|\frac{1+t}{2+t}\right|+\mathrm{C}=\log \left(\frac{1+e^{x}}{2+e^{x}}\right)+\mathrm{C}\)

Question 14.
\(\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}\)
Solution:

Question 15.
\(\cos ^{3} x e^{\log \sin x}\)
Solution:

Question 16.
\(e^{3 \log x}\left(x^{4}+1\right)^{-1}\)
Solution:

Question 17.
\(f^{\prime}(a x+b)[f(a x+b)]^{n}\)
Solution:

Question 18.
\(\frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}\)
Solution:

Question 19.
\(\frac{\sin ^{-1} \sqrt{x}-\cos ^{-1} \sqrt{x}}{\sin ^{-1} \sqrt{x}+\cos ^{-1} \sqrt{x}}, x \in[0,1]\)
Solution:

Question 20.
\(\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\)
Solution:

Question 21.
\(\frac{2+\sin 2 x}{1+\cos 2 x} e^{x}\)
Solution:

Question 22.
\(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}\)
Solution:
Let \(\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{\mathrm{A}}{(x+1)}+\frac{\mathrm{B}}{(x+1)^{2}}+\frac{\mathrm{C}}{(x+2)}\)
⇒ x² + x + 1 = A(x + 1)(x + 2) + B(x + 2) + C(x + 1)² … (1)
Put x = – 1 in (1), we get B = 1
Put x = – 2 in (1), we get C = 3
Equating the coefficients of x2, we get
A + C = 1 ∴ A = – 2

Question 23.
\(\tan ^{-1} \sqrt{\frac{1-x}{1+x}}\)
Solution:

Question 24.
\(\frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}\)
Solution:

Question 25.
\(\int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x\)
Solution:

Question 26.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x\)
Solution:

Question 27.
\(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}\)
Solution:
Let I = \(\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}\)
Dividing the Nr. and Dr. by cos²x, we get

where t² = y
⇒ 1 = A(1 + 4y) + B(1 + y) … (1)
Put y = – 1 in (1), we get 1 = – 3A ∴ A = \(\frac { – 1 }{ 3 }\)
Equating the coefficients of y, we get
4A + B = 0
∴ B = \(\frac { 4 }{ 3 }\)

Question 28.
\(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Solution:

Question 29.
\(\int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}\)
Solution:

Question 30.
\(\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x\)
Solution:

Question 31.
\(\int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x\)
Solution:

Question 32.
\(\int_{0}^{\pi} \frac{x \tan x}{\sec x+\tan x} d x\)
Solution:

Question 33.
\(\left.\int_{1}^{4}|| x-1|+| x-2|+| x-3 \mid\right] d x\)
Solution:

Question 34.
\(\int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}\)
Solution:
Let \(\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1}\)
1 = Ax(x + 1) + B(x + 1) + C(x²) … (1)
Put x = 0 in (1), we get B = 1
Put x = – 1 in (1), we get C = 1
Equating the coefficients of x², we get
A + C = 0 ∴ A = – 1

Question 35.
\(\int_{0}^{1} x e^{x} d x=1\)
Solution:

Question 36.
\(\int_{-1}^{1} x^{17} \cos ^{4} x d x=0\)
Solution:
Let f(x) = x¹⁷cos⁴x
f(- x) = (- x)¹⁷cos< sup>4(-x)
∴ f(x) is an odd function.
∴ \(\int_{-1}^{1} x^{17} \cos ^{4} x \cdot d x=0\)
\(\int_{-a}^{a} f(x) d x=0, \text { if } f(x) \text { is odd }\)

Question 37.
\(\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}\)
Solution:

Question 38.
\(\int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2\)
Solution:

Question 39.
\(\int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1\)
Solution
∫sin^-1 x dx = x sin^-1 x + \(\sqrt{1-x^{2}}\)
(Refer in-tegrals of inverse trigonometric functions)
∴ \(\int_{0}^{1} \sin ^{-1} x d x=\left[x \sin ^{-1} x+\sqrt{1-x^{2}}\right]_{0}^{1}\)
= \(\left(\frac{\pi}{2}+0\right)-(0+1)=\frac{\pi}{2}-1\)

Question 40.
Evaluate \(\int_{0}^{1} e^{2-3 x} d x\) as a limit of a sum.
Solution:
Let I = \(\int_{0}^{1} e^{2-3 x} d x\)
Here f(x) = e^2-3x, a = 0, b = 1
nh = b – a = 1 – 0 = 1
f(0 + h) = f(h) = e^2-3h
f(0 + 2h) = f(2h) = e^2-6h
……………………………….
f(0 + (n – 1)h = f((n – 1)h) = e^2-3(n-1)h

Question 41.
\(\int \frac{d x}{e^{x}+e^{-x}}\) is equal to
a. tan^-1(e^x) + C
b. tan^-1(e^-x) + C
c. log(e^x – e^-x) + C
d. log(e^x + x^-x) + C
Solution:

Question 42.
\(\int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x\) is equal to
a. \(\frac{-1}{\sin x+\cos x}+\mathrm{C}\)
b. log|sin x + cos x| + C
c. log|sin x – cos x| + C
d. \(\frac{1}{(\sin x+\cos x)^{2}}\)
Solution:

Question 43.
If \(f(a+b-x)=f(x), \text { then } \int_{a}^{b} x f(x) d x\) is equal to
a. \(\frac{a+b}{2} \int_{a}^{b} f(b-x) d x\)
b. \(\frac{a+b}{2} \int_{a}^{b} f(b+x) d x\)
c. \(\frac{b-a}{2} \int_{a}^{b} f(x) d x\)
d. \(\frac{a+b}{2} \int_{a}^{b} f(x) d x\)
Solution:

Question 44.
The value of \(\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x\) is
a. 1
b. 0
c. – 1
d. \(\frac{\pi}{4}\)
Solution: