CBSE Class 12 Mathematics Integrals Important Questions Set 02

Question. Evaluate: \(\int_{0}^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\) 
Answer: Let \(I = \int_{0}^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\)
\(\Rightarrow I = \int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2}-x\right) \sin\left(\frac{\pi}{2}-x\right) \cos\left(\frac{\pi}{2}-x\right)}{\sin^4\left(\frac{\pi}{2}-x\right) + \cos^4\left(\frac{\pi}{2}-x\right)} dx\) [By Property \(\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx\)]
\(\Rightarrow I = \int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2}-x\right) \cos x \sin x}{\cos^4 x + \sin^4 x} dx\) \([\because \sin\left(\frac{\pi}{2}-x\right) = \cos x \text{ and } \cos\left(\frac{\pi}{2}-x\right) = \sin x]\)
\(\Rightarrow I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\cos x \sin x}{\sin^4 x + \cos^4 x} dx - \int_{0}^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\)
\(\Rightarrow I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\cos x \sin x}{\sin^4 x + \cos^4 x} dx - I \Rightarrow 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx\)
\(= \frac{\pi}{2} \int_{0}^{\pi/2} \frac{\frac{\sin x \cos x}{\cos^4 x}}{\frac{\sin^4 x + \cos^4 x}{\cos^4 x}} dx\) [Dividing numerator and denominator by \(\cos^4 x\)]
\(= \frac{\pi}{2 \times 2} \int_{0}^{\pi/2} \frac{2 \tan x \cdot \sec^2 x}{1 + (\tan^2 x)^2} dx\)
Let \(\tan^2 x = z\); \(2 \tan x \cdot \sec^2 x dx = dz\)
The limits are, when \(x = 0, z = 0; x = \frac{\pi}{2}, z = \infty\)
\(\therefore 2I = \frac{\pi}{4} \int_{0}^{\infty} \frac{dz}{1+z^2} = \frac{\pi}{4} [\tan^{-1} z]_0^{\infty} = \frac{\pi}{4} (\tan^{-1} \infty - \tan^{-1} 0)\)
\(\therefore 2I = \frac{\pi}{4} \left(\frac{\pi}{2} - 0\right) \Rightarrow I = \frac{\pi^2}{16}\)

Question. Evaluate: \(\int_{0}^{\pi/2} 2 \sin x \cos x \tan^{-1}(\sin x) dx\) 
Answer: Let \(I = \int_{0}^{\pi/2} 2 \sin x \cos x \tan^{-1}(\sin x) dx\)
Put \(\sin x = z \Rightarrow \cos x dx = dz\)
The limits are, when \(x = 0, z = \sin 0 = 0; x = \frac{\pi}{2}, z = \sin \frac{\pi}{2} = 1\)
\(\therefore I = \int_{0}^{1} 2z \tan^{-1}(z) dz = 2 \left[ \tan^{-1} z \cdot \frac{z^2}{2} \right]_0^1 - 2 \int_{0}^{1} \frac{1}{1+z^2} \cdot \frac{z^2}{2} dz\)
\(= 2 \left[ \frac{\pi}{4} \cdot \frac{1}{2} - 0 \right] - \int_{0}^{1} \frac{z^2}{1+z^2} dz = \frac{\pi}{4} - \int_{0}^{1} \frac{1+z^2-1}{1+z^2} dz = \frac{\pi}{4} - \int_{0}^{1} dz + \int_{0}^{1} \frac{dz}{1+z^2}\)
\(= \frac{\pi}{4} - [z]_0^1 + [\tan^{-1} z]_0^1 = \frac{\pi}{4} - 1 + \left[\frac{\pi}{4} - 0\right] = \frac{\pi}{2} - 1\)

Question. Evaluate: \(\int_{0}^{\pi} \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\)
Answer: Let \(I = \int_{0}^{\pi} \frac{x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\) ...(i)
\(I = \int_{0}^{\pi} \frac{\pi - x}{a^2 \cos^2(\pi - x) + b^2 \sin^2(\pi - x)} dx\) [\(\because \int_0^a f(x) dx = \int_0^a f(a-x) dx\)]
\(I = \int_{0}^{\pi} \frac{\pi - x}{a^2 \cos^2 x + b^2 \sin^2 x} dx\) ...(ii)
Adding (i) and (ii), we get
\(2I = \int_{0}^{\pi} \frac{\pi}{a^2 \cos^2 x + b^2 \sin^2 x} dx \Rightarrow I = \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{a^2 \cos^2 x + b^2 \sin^2 x}\)
\(I = \frac{\pi}{2} \int_{0}^{\pi} \frac{\sec^2 x}{a^2 + b^2 \tan^2 x} dx\) [Divide numerator and denominator by \(\cos^2 x\)]
\(= \pi \int_{0}^{\pi/2} \frac{\sec^2 x}{a^2 + b^2 \tan^2 x} dx\) [\(\because \int_0^{2a} f(x) dx = 2 \int_0^a f(x) dx\)]
Put \(b \tan x = t \Rightarrow b \sec^2 x dx = dt\)
The limits are, when \(x = 0, t = 0\) and \(x = \frac{\pi}{2}, t = \infty\)
\(I = \frac{\pi}{b} \int_{0}^{\infty} \frac{dt}{a^2 + t^2} = \frac{\pi}{b} \cdot \frac{1}{a} \left[ \tan^{-1} \frac{t}{a} \right]_0^{\infty}\)
\(I = \frac{\pi}{ab} (\tan^{-1} \infty - \tan^{-1} 0) = \frac{\pi}{ab} \cdot \frac{\pi}{2} \Rightarrow I = \frac{\pi^2}{2ab}\)

Question. Solve: \(\int \sqrt{\frac{a+x}{a-x}} dx\) 
Answer: Let \(I = \int \sqrt{\frac{a+x}{a-x}} dx\)
Put \(x = a \cos 2\theta\)
\(\Rightarrow dx = -a \cdot \sin 2\theta \cdot 2 \cdot d\theta\)
\(\therefore I = -2 \int \sqrt{\frac{a + a \cos 2\theta}{a - a \cos 2\theta}} \cdot a \sin 2\theta d\theta\) \([\because \cos 2\theta = \frac{x}{a} \Rightarrow 2\theta = \cos^{-1} \frac{x}{a} \Rightarrow \theta = \frac{1}{2} \cos^{-1} \frac{x}{a}]\)
\(= -2a \int \sqrt{\frac{1 + \cos 2\theta}{1 - \cos 2\theta}} \sin 2\theta d\theta = -2a \int \sqrt{\frac{2 \cos^2 \theta}{2 \sin^2 \theta}} \sin 2\theta d\theta\)
\(= -2a \int \cot \theta \cdot \sin 2\theta d\theta = -2a \int \frac{\cos \theta}{\sin \theta} \cdot 2 \sin \theta \cdot \cos \theta d\theta = -4a \int \cos^2 \theta d\theta = -2a \int (1 + \cos 2\theta) d\theta\)
\(= -2a \left[ \theta + \frac{\sin 2\theta}{2} \right] + C = -2a \left[ \frac{1}{2} \cos^{-1} \frac{x}{a} + \frac{1}{2} \sqrt{1 - \frac{x^2}{a^2}} \right] + C\)
\(= -a \left[ \cos^{-1}\left(\frac{x}{a}\right) + \sqrt{1 - \frac{x^2}{a^2}} \right] + C\)

Question. Evaluate the following: \(\int_{0}^{3/2} |x \cos \pi x| dx\) 
Answer: Sol. \(\int_{0}^{3/2} |x \cos \pi x| dx\)
As we know, \(\cos x = 0 \Rightarrow x = (2n-1)\frac{\pi}{2}, n \in Z\)
\(\therefore \cos \pi x = 0 \Rightarrow x = \frac{1}{2}, \frac{3}{2}\)
For \(0 < x < \frac{1}{2}\), \(x > 0\) then \(\cos \pi x > 0 \Rightarrow x \cos \pi x > 0\)
For \(\frac{1}{2} < x < \frac{3}{2}\), \(x > 0\) then \(\cos \pi x < 0 \Rightarrow x \cos \pi x < 0\)
\(\therefore \int_{0}^{3/2} |x \cos \pi x| dx = \int_{0}^{1/2} x \cos \pi x dx + \int_{1/2}^{3/2} (-x \cos \pi x) dx\) ...(i)
\(= \left[ x \frac{\sin \pi x}{\pi} \right]_0^{1/2} - \int_{0}^{1/2} 1 \cdot \frac{\sin \pi x}{\pi} dx - \left[ x \frac{\sin \pi x}{\pi} \right]_{1/2}^{3/2} + \int_{1/2}^{3/2} \frac{\sin \pi x}{\pi} dx\)
\(= \left[ \frac{x}{\pi} \sin \pi x + \frac{1}{\pi^2} \cos \pi x \right]_0^{1/2} - \left[ \frac{x}{\pi} \sin \pi x + \frac{1}{\pi^2} \cos \pi x \right]_{1/2}^{3/2}\)
\(= \left( \frac{1}{2\pi} + 0 - \frac{1}{\pi^2} \right) - \left( -\frac{3}{2\pi} - \frac{1}{2\pi} \right) = \frac{5}{2\pi} - \frac{1}{\pi^2}\)

Question. Evaluate: \(\int_{0}^{\pi} \frac{x}{1 + \sin \alpha \sin x} dx\)
Answer: Let \(I = \int_{0}^{\pi} \frac{x}{1 + \sin \alpha \sin x} dx = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin \alpha \sin(\pi - x)} dx\)
\(= \int_{0}^{\pi} \frac{\pi}{1 + \sin \alpha \sin x} dx - \int_{0}^{\pi} \frac{x}{1 + \sin \alpha \sin x} dx\)
\(I = \pi \int_{0}^{\pi} \frac{dx}{1 + \sin \alpha \sin x} - I\)
\(\Rightarrow 2I = \pi \int_{0}^{\pi} \frac{dx}{1 + \sin \alpha \sin x} = \pi \int_{0}^{\pi} \frac{dx}{1 + \sin \alpha \cdot \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}}\)
\(= \pi \int_{0}^{\pi} \frac{(1 + \tan^2 \frac{x}{2})}{1 + \tan^2 \frac{x}{2} + 2 \sin \alpha \tan \frac{x}{2}} dx = \pi \int_{0}^{\pi} \frac{\sec^2 \frac{x}{2}}{\tan^2 \frac{x}{2} + 2 \sin \alpha \tan \frac{x}{2} + 1} dx\)
Let \(\tan \frac{x}{2} = t \Rightarrow \sec^2 \frac{x}{2} dx = 2dt; x = 0 \Rightarrow t = 0 \text{ and } x = \pi \Rightarrow t = \infty\)
\(\therefore 2I = 2\pi \int_{0}^{\infty} \frac{dt}{t^2 + 2 \sin \alpha t + 1}\)
\(I = \pi \int_{0}^{\infty} \frac{dt}{t^2 + 2 \sin \alpha t + \sin^2 \alpha - \sin^2 \alpha + 1}\)
\(= \pi \int_{0}^{\infty} \frac{dt}{(t + \sin \alpha)^2 + (1 - \sin^2 \alpha)} = \pi \int_{0}^{\infty} \frac{dt}{(t + \sin \alpha)^2 + \cos^2 \alpha}\)
\(= \frac{\pi}{\cos \alpha} \left[ \tan^{-1} \frac{t + \sin \alpha}{\cos \alpha} \right]_0^{\infty} = \frac{\pi}{\cos \alpha} \left[ \tan^{-1} \left( \frac{\tan \frac{x}{2} + \sin \alpha}{\cos \alpha} \right) \right]_0^{\infty}\)
\(= \frac{\pi}{\cos \alpha} \left[ \frac{\pi}{2} - \tan^{-1}(\tan \alpha) \right] = \frac{\pi}{\cos \alpha} \left( \frac{\pi}{2} - \alpha \right)\)
\(= \frac{\pi(\pi - 2\alpha)}{2 \cos \alpha}\)

Question. Find: \(\int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx\)
Answer: Let \(I = \int \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}} dx\)
Putting \(\sqrt{x} = \cos \theta\), i.e., \(x = \cos^2 \theta \Rightarrow dx = -2 \cos \theta \sin \theta d\theta\), we get
\(I = \int \sqrt{\frac{1-\cos \theta}{1+\cos \theta}} (-2 \sin \theta \cos \theta) d\theta\)
\(= -2 \int \sqrt{\frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}}} (\sin \theta \cos \theta) d\theta = -2 \int \frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}} (2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cos \theta) d\theta\)
\(= -2 \int 2 \sin^2 \frac{\theta}{2} \cos \theta d\theta = -2 \int (1 - \cos \theta) \cos \theta d\theta\)
\(= -2 \int (1 - \cos \theta) \cos \theta \cdot d\theta = -2 \int (\cos \theta - \cos^2 \theta) \cdot d\theta\)
\(= -2 \int \cos \theta \cdot d\theta + \int 2 \cos^2 \theta \cdot d\theta = -2 \sin \theta + \int (1 + \cos 2\theta) \cdot d\theta\)
\(= -2 \sin \theta + \theta + \frac{\sin 2\theta}{2} + C = -2 \sin \theta + \theta + \frac{2 \sin \theta \cos \theta}{2} + C\)
\(= -2 \sin \theta + \theta + \sin \theta \cos \theta + C\)
\(= -2\sqrt{1-\cos^2 \theta} + \theta + \sqrt{1-\cos^2 \theta} \cdot \cos \theta + C\)
\(= -2\sqrt{1-x} + \cos^{-1}\sqrt{x} + \sqrt{x}\sqrt{1-x} + C\)

Question. Find: \(\int \frac{x^2}{(x \sin x + \cos x)^2} dx\)
Answer: Let \(I = \int \frac{x^2}{(x \sin x + \cos x)^2} dx = \int \frac{x}{\cos x} \cdot \frac{x \cos x}{(x \sin x + \cos x)^2} dx\)
Integrating by parts, taking \(\frac{x}{\cos x}\) as the first function and \(\frac{x \cos x}{(x \sin x + \cos x)^2}\) as the second function, we get
\(I = \frac{x}{\cos x} \cdot \int \frac{x \cos x}{(x \sin x + \cos x)^2} dx - \int \left[ \frac{d}{dx}\left(\frac{x}{\cos x}\right) \int \left(\frac{x \cos x}{(x \sin x + \cos x)^2}\right) dx \right] dx\)
Now, let us first evaluate \(\int \frac{x \cos x dx}{(x \sin x + \cos x)^2}\)
Putting \((x \sin x + \cos x) = t\), then \((\sin x + x \cos x - \sin x)dx = dt\) i.e., \(x \cos x dx = dt\), we get
\(\int \frac{x \cos x}{(x \sin x + \cos x)^2} dx = \int \frac{dt}{t^2} = -\frac{1}{t} = -\frac{1}{(x \sin x + \cos x)}\)
Hence, \(I = \frac{x}{\cos x} \cdot \frac{-1}{(x \sin x + \cos x)} - \int \frac{\cos x + x \sin x}{\cos^2 x} \times \frac{-1}{(x \sin x + \cos x)} dx\)
\(= \frac{-x}{\cos x (x \sin x + \cos x)} + \int \sec^2 x dx = \frac{-x}{\cos x (x \sin x + \cos x)} + \tan x + C\)
\(= \frac{-x + \sin x (x \sin x + \cos x)}{\cos x (x \sin x + \cos x)} + C = \frac{-x + x \sin^2 x + \sin x \cos x}{\cos x (x \sin x + \cos x)} + C\)
\(= \frac{-x(1 - \sin^2 x) + \sin x \cos x}{\cos x (x \sin x + \cos x)} + C\)
\(= \frac{-x \cos^2 x + \sin x \cos x}{\cos x (x \sin x + \cos x)} + C = \frac{\sin x - x \cos x}{x \sin x + \cos x} + C\)
\(\int \frac{x^2 dx}{(x \sin x + \cos x)^2} = \frac{(\sin x - x \cos x)}{(x \sin x + \cos x)} + C\)

Objective Type Questions:

Question. \(\int \frac{x^9}{(4x^2 + 1)^6} dx\) is equal to 
(a) \(\frac{1}{5x}(4 + \frac{1}{x^2})^{-5} + C\)
(b) \(\frac{1}{5}(4 + \frac{1}{x^2})^{-5} + C\)
(c) \(\frac{1}{10x}(1 + 4)^{-5} + C\)
(d) \(\frac{1}{10}(\frac{1}{x^2} + 4)^{-5} + C\)
Answer: (d)

Question. The integral of \(\int \frac{x}{\sqrt{x + 1}} dx\) is equal to
(a) \(2\left[\frac{x\sqrt{x}}{3} - \frac{x}{2} + \sqrt{x} - \log|(\sqrt{x} + 1)|\right] + C\)
(b) \(\frac{x\sqrt{x}}{3} + \frac{x}{2} - \sqrt{x} + \log(\sqrt{x} + 1) + C\)
(c) \(\sqrt{x} - \log(\sqrt{x} + 1) + C\)
(d) None of these
Answer: (a)

Question. \(\int e^x [f(x) + f'(x)] dx = e^x \sin x + C\) then \(f(x)\) is equal to
(a) \(\sin x\)
(b) \(-\sin x\)
(c) \(\cos x - \sin x\)
(d) \(\sin x + \cos x\)
Answer: (a)

Question. \(\int \frac{a^{\sqrt{x}}}{\sqrt{x}} dx\) is
(a) \(a^{\sqrt{x}} \log a + C\)
(b) \(2a^{\sqrt{x}} \log_e a + C\)
(c) \(2a^{\sqrt{x}} \log_{10} a + C\)
(d) \(\frac{2a^{\sqrt{x}}}{\log_e a} + C\)
Answer: (d)

Question. \(\int_1^3 \frac{3 \cos(\log x)}{x} dx\) is equal to
(a) \(\sin(\log 3)\)
(b) \(\cos(\log 3)\)
(c) 1
(d) \(\frac{\pi}{4}\)
Answer: (a)

Question. \(\int_0^1 \frac{\tan^{-1} x}{1 + x^2} dx\) is equal to
(a) 1
(b) \(\frac{\pi^2}{4}\)
(c) \(\frac{\pi^2}{32}\)
(d) none of these
Answer: (c)

Fill in the blanks.

Question. \(\int \frac{\sin x}{3 + 4 \cos^2 x} dx = \) _____________ . 
Answer: \(-\frac{1}{2\sqrt{3}} \tan^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right) + C\)

Question. \(\int \frac{2 dx}{\sqrt{1 - 4x^2}} = \) _____________ .
Answer: \(\sin^{-1}(2x) + C\)

Question. \(\int_{-a}^a f(x) dx = 0\) if \(f(x)\) is an _____________ function. 
Answer: odd

Question. \(\int_3^6 2[x] dx = \) _____________ , where \([x]\) is the greatest integer function.
Answer: 24

Very Short Answer Questions:

Question. If \(\int (e^{ax} + bx) dx = \frac{e^{4x}}{4} + \frac{3x^2}{2}\), find the values of \(a\) and \(b\).
Answer: \(a = 4, b = 3\)

Question. If \(\int_0^a 3x^2 dx = 8\), write the value of 'a'. 
Answer: \(a = 2\)

Question. If \(\int_0^1 (3x^2 + 2x + k) dx = 0\), find the value of \(k\). 
Answer: \(k = -2\)

Question. If \(f(x) = \int_0^x t \sin t dt\), then write the value of \(f'(x)\). 
Answer: \(x \sin x\)

Question. Write the antiderivative of \(\left(3\sqrt{x} + \frac{1}{\sqrt{x}}\right)\).
Answer: \(2x^{3/2} + 2\sqrt{x} + C\)

Question. If \(\int (ax + b)^2 dx = f(x) + C\), find \(f(x)\).
Answer: \frac{(ax+b)^3}{3a}

Question. Evaluate: \(\int_0^1 \frac{1}{\sqrt{2x + 3}} dx\) 
Answer: \(\sqrt{5} - \sqrt{3}\)

Question. Evaluate: \(\int \frac{dx}{\sin^2 x \cos^2 x}\) 
Answer: \(\tan x - \frac{1}{\tan x} + C\)

Question. Evaluate: \(\int_0^{\pi/4} \tan x dx\) 
Answer: \(\frac{1}{2} \log 2\)

Question. Evaluate: \(\int \cos^{-1}(\sin x) dx\) 
Answer: \(\frac{\pi x}{2} - \frac{x^2}{2} + C\)

Question. Evaluate : \(\int_e^{e^2} \frac{dx}{x \log x}\)
Answer: \(\log 2\)

Question. Evaluate: \(\int \frac{dx}{\sqrt{1 - x^2}}\)
Answer: \(\sin^{-1} x + C\)

Question. Write the value of \(\int \frac{dx}{x^2 + 16}\).
Answer: \(\frac{1}{4} \tan^{-1} \frac{x}{4} + C\)

Short Answer Questions–:

Question. Given \(\int e^x (\tan x + 1) \sec x dx = e^x f(x) + C\). Write \(f(x)\) satisfying the above. 
Answer: \(f(x) = \sec x\)

Question. Evaluate: \(\int (1 - x)\sqrt{x} dx\) 
Answer: \(\frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C\)

Question. If \(\int \left(\frac{x - 1}{x^2}\right) e^x dx = f(x) e^x + C\), find the value of \(f(x)\). 
Answer: \(f(x) = \frac{1}{x}\)

Question. Evaluate : \(\int_0^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx\) 
Answer: \(\frac{\pi^2}{4}\)

Question. Show that \(\int_0^{\pi/2} (\sqrt{\tan x} + \sqrt{\cot x}) dx = \sqrt{2}\pi\). 
Answer: \(\sqrt{2}\pi\) (As shown in proof)

Question. If \(\int_0^a \frac{1}{4 + x^2} dx = \frac{\pi}{8}\), find the value of \(a\). 
Answer: \(a = 2\)

Question. Evaluate : \(\int_0^{\pi/4} \left(\frac{\sin x + \cos x}{3 + \sin 2x}\right) dx\) 
Answer: \(\frac{1}{4} \log 3\)

Question. Find : \(\int \frac{dx}{1 + \tan x}\) 
Answer: \(\frac{x}{2} + \frac{1}{2} \log |\cos x + \sin x| + C\)

Question. Evaluate : \(\int_0^{\pi/2} \left(\frac{5 \sin x + 3 \cos x}{\sin x + \cos x}\right) dx\) 
Answer: \(2\pi\)

Question. Evaluate : \(\int \frac{(x + 3)e^x}{(x + 5)^3} dx\)
Answer: \(\frac{e^x}{(x+5)^2} + C\)

Question. Evaluate : \(\int_0^{\pi/2} \frac{2^{\sin x}}{2^{\sin x} + 2^{\cos x}} dx\) 
Answer: \(\frac{\pi}{4}\)

Question. Evaluate : \(\int_0^{\pi/2} e^x (\sin x - \cos x) dx\) 
Answer: 1

Question. Evaluate: \(\int \frac{\cos \sqrt{x}}{\sqrt{x}} dx\) 
Answer: \(2 \sin \sqrt{x} + C\)

Question. Write the value of the following integral \(\int_{-\pi/2}^{\pi/2} \sin^5 x dx\). 
Answer: 0

Question. Evaluate: \(\int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} dx\) 
Answer: \(x - \sqrt{1 - x^2} \sin^{-1} x + C\)