Match List-I with List-II
| List-I | List-II | ||
|---|---|---|---|
| Definite integral | Value | ||
| (A) | \(\int_{0}^{\frac{\pi}{2}}\displaystyle\sin^{2}\frac{x}{2}\,dx\) | (I) | \(\sqrt{2}-1\) |
| (B) | \(\int_{0}^{\frac{\pi}{4}}\displaystyle\frac{\cos 2x}{\cos x+\sin x}\,dx\) | (II) | \(\displaystyle\frac{\pi}{4}-\frac{1}{2}\) |
| (C) | \(\int_{0}^{\frac{\pi}{4}}\displaystyle\frac{1-\tan^{2}x}{1+\tan^{2}x}\,dx\) | (III) | \(\displaystyle\frac{\pi}{2}+1\) |
| (D) | \(\int_{0}^{\frac{\pi}{2}}\displaystyle\left(\sin x+\cos x\right)^{2}dx\) | (IV) | \(\displaystyle\frac{1}{2}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(A) \(\int_{0}^{\pi/2}\sin^{2}\frac{x}{2}\,dx\)
This matches (II) \(\frac{\pi}{4}-\frac{1}{2}\), since \(\sin^{2}\frac{x}{2} = \frac{1-\cos x}{2}\):
\( \displaystyle \left[\frac{x}{2} - \frac{\sin x}{2}\right]_{0}^{\pi/2} = \frac{\pi}{4} - \frac{1}{2} \)
(B) \(\int_{0}^{\pi/4}\frac{\cos 2x}{\cos x+\sin x}\,dx\)
This matches (I) \(\sqrt{2}-1\), since \(\cos 2x = (\cos x-\sin x)(\cos x+\sin x)\) leaves \(\cos x - \sin x\):
\( \displaystyle \left[\sin x + \cos x\right]_{0}^{\pi/4} = \left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) - 1 = \sqrt{2} - 1 \)
(C) \(\int_{0}^{\pi/4}\frac{1-\tan^{2}x}{1+\tan^{2}x}\,dx\)
This matches (IV) \(\frac{1}{2}\), since \(\frac{1-\tan^{2}x}{1+\tan^{2}x} = \cos 2x\):
\( \displaystyle \left[\frac{\sin 2x}{2}\right]_{0}^{\pi/4} = \frac{1}{2} \)
(D) \(\int_{0}^{\pi/2}\left(\sin x+\cos x\right)^{2}dx\)
This matches (III) \(\frac{\pi}{2}+1\), since \((\sin x+\cos x)^{2} = 1 + \sin 2x\):
\( \displaystyle \left[x - \frac{\cos 2x}{2}\right]_{0}^{\pi/2} = \left(\frac{\pi}{2} + \frac{1}{2}\right) - \left(0 - \frac{1}{2}\right) = \frac{\pi}{2} + 1 \)