Match List I with List II
| LIST I | LIST II | ||
| A. | \(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\sin^{\frac{7}{2}}x}{\sin^{\frac{7}{2}}x+\cos^{\frac{7}{2}}x}\,dx\) | I. | \(\displaystyle\frac{\pi}{4}-\frac{1}{2}\) |
| B. | \(\int_{0}^{\pi}\displaystyle\frac{x\sin x}{1+\cos^2 x}\,dx\) | II. | 0 |
| C. | \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\displaystyle x\cos x\,dx\) | III. | \(\displaystyle\frac{\pi}{4}\) |
| D. | \(\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\displaystyle\sin^2 x\,dx\) | IV. | \(\displaystyle\frac{\pi^2}{4}\) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-IV, C-II, D-I
(A) \(\displaystyle\int_{0}^{\pi/2}\frac{\sin^{7/2}x}{\sin^{7/2}x+\cos^{7/2}x}\,dx\)
This matches (III) \(\frac{\pi}{4}\).
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 and cos change places:
\( \displaystyle I = \int_{0}^{\pi/2}\frac{\cos^{7/2}x}{\cos^{7/2}x+\sin^{7/2}x}\,dx \)
Adding the two forms of I, the integrand becomes 1:
\( \displaystyle 2I = \int_{0}^{\pi/2} 1\,dx = \frac{\pi}{2} \)
\( \displaystyle I = \frac{\pi}{4} \)
(B) \(\displaystyle\int_{0}^{\pi}\frac{x\sin x}{1+\cos^2 x}\,dx\)
This matches (IV) \(\frac{\pi^2}{4}\).
Call the integral I. Replacing x by \(\pi - x\) (sin x stays the same, \(\cos^2 x\) stays the same):
\( \displaystyle I = \int_{0}^{\pi}\frac{(\pi - x)\sin x}{1+\cos^2 x}\,dx \)
Adding the two forms:
\( \displaystyle 2I = \pi\int_{0}^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx \)
Put t = cos x, dt = −sin x dx; x = 0 gives t = 1 and x = π gives t = −1:
\( \displaystyle 2I = \pi\int_{-1}^{1}\frac{dt}{1+t^2} = \pi\left[\tan^{-1}t\right]_{-1}^{1} \)
\( \displaystyle 2I = \pi\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \frac{\pi^2}{2} \)
\( \displaystyle I = \frac{\pi^2}{4} \)
(C) \(\displaystyle\int_{-\pi/2}^{\pi/2} x\cos x\,dx\)
This matches (II) 0.
Here \(f(x) = x\cos x\) is an odd function:
\(f(-x) = (-x)\cos(-x) = -x\cos x = -f(x)\)
The integral of an odd function over \([-a, a]\) is 0.
(D) \(\displaystyle\int_{-\pi/4}^{\pi/4}\sin^2 x\,dx\)
This matches (I) \(\frac{\pi}{4}-\frac{1}{2}\).
\(\sin^2 x\) is an even function, so the integral is twice the integral over \([0, \frac{\pi}{4}]\). Using \(2\sin^2 x = 1 - \cos 2x\):
\( \displaystyle \int_{-\pi/4}^{\pi/4}\sin^2 x\,dx = \int_{0}^{\pi/4} 2\sin^2 x\,dx = \int_{0}^{\pi/4}(1 - \cos 2x)\,dx \)
\( \displaystyle = \left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi/4} \)
\( \displaystyle = \frac{\pi}{4} - \frac{1}{2} \)