Match List - I with List - II. Evaluate the integrals.
| List - I | List - II | ||
| (A) | \(\int_{0}^{\pi/2}\displaystyle\frac{(\sin x-\cos x)\,dx}{1+\sin x\cos x}\) | (I) | \(2\pi\) |
| (B) | \(\int_{0}^{1/2}\displaystyle\frac{dx}{\sqrt{x-x^2}}\) | (II) | \(\displaystyle\frac{\pi}{4}\) |
| (C) | \(\int_{-\pi}^{\pi}x\sin x\,dx\) | (III) | 0 |
| (D) | \(\int_{0}^{1}\displaystyle\frac{1}{1+x^2}\,dx\) | (IV) | \(\displaystyle\frac{\pi}{2}\) |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
(A) \(\displaystyle\int_{0}^{\pi/2}\frac{(\sin x-\cos x)\,dx}{1+\sin x\cos x}\)
This matches (III) 0.
Call the integral I. By the property \(\int_{0}^{a}f(x)\,dx = \int_{0}^{a}f(a-x)\,dx\), replace x by \(\frac{\pi}{2}-x\): sin x and cos x change places, so the numerator changes sign and the denominator stays the same.
\( \displaystyle I = \int_{0}^{\pi/2}\frac{(\cos x-\sin x)\,dx}{1+\cos x\sin x} = -I \)
\(2I = 0,\quad I = 0\)
(B) \(\displaystyle\int_{0}^{1/2}\frac{dx}{\sqrt{x-x^2}}\)
This matches (IV) \(\frac{\pi}{2}\).
Put \(x = \sin^2 t\), so \(dx = 2\sin t\cos t\,dt\) and \(\sqrt{x - x^2} = \sqrt{\sin^2 t\cos^2 t} = \sin t\cos t\). When x = 0, t = 0; when \(x = \frac{1}{2}\), \(\sin t = \frac{1}{\sqrt{2}}\), so \(t = \frac{\pi}{4}\).
\( \displaystyle \int_{0}^{1/2}\frac{dx}{\sqrt{x-x^2}} = \int_{0}^{\pi/4}\frac{2\sin t\cos t}{\sin t\cos t}\,dt \)
\( \displaystyle = \int_{0}^{\pi/4}2\,dt = \frac{\pi}{2} \)
(C) \(\displaystyle\int_{-\pi}^{\pi}x\sin x\,dx\)
This matches (I) \(2\pi\).
x and sin x are both odd, so x sin x is an even function and
\( \displaystyle \int_{-\pi}^{\pi}x\sin x\,dx = 2\int_{0}^{\pi}x\sin x\,dx \)
Integrating by parts, \(\int x\sin x\,dx = -x\cos x + \sin x\):
\( \displaystyle 2\Big[-x\cos x + \sin x\Big]_{0}^{\pi} = 2\big[(-\pi)(-1) + 0\big] - 0 \)
\(= 2\pi\)
(D) \(\displaystyle\int_{0}^{1}\frac{1}{1+x^2}\,dx\)
This matches (II) \(\frac{\pi}{4}\).
\( \displaystyle \int_{0}^{1}\frac{dx}{1+x^2} = \Big[\tan^{-1}x\Big]_{0}^{1} = \frac{\pi}{4} - 0 = \frac{\pi}{4} \)