The value of \(\int_{0}^{\frac{\pi}{2}}\displaystyle\frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x}\,dx\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{\pi}{4}\)
Since \(\sin^{4}x+\cos^{4}x = \left(\sin^{2}x+\cos^{2}x\right)^{2} - 2\sin^{2}x\cos^{2}x\),
\( \displaystyle \sin^{4}x+\cos^{4}x = 1 - \frac{1}{2}\sin^{2}2x = \frac{1+\cos^{2}2x}{2} \)
Also \(\sin x\cos x = \frac{1}{2}\sin 2x\). So the integrand is
\( \displaystyle \frac{\sin x\cos x}{\sin^{4}x+\cos^{4}x} = \frac{\sin 2x}{1+\cos^{2}2x} \)
Let \(t = \cos 2x\). Then \(dt = -2\sin 2x\,dx\).
When \(x = 0\), \(t = 1\); when \(x = \frac{\pi}{2}\), \(t = -1\). So
\( \displaystyle \int_{0}^{\pi/2}\frac{\sin 2x}{1+\cos^{2}2x}\,dx = -\frac{1}{2}\int_{1}^{-1}\frac{dt}{1+t^{2}} \)
\( \displaystyle = \frac{1}{2}\left[\tan^{-1}t\right]_{-1}^{1} = \frac{1}{2}\left(\frac{\pi}{4} + \frac{\pi}{4}\right) \)
\( \displaystyle = \frac{\pi}{4} \)
Why the other options are incorrect
Option (2) \(\displaystyle\frac{\pi}{2}\) — Incorrect. It drops the factor \(\frac{1}{2}\) that comes from \(dt = -2\sin 2x\,dx\).
Option (3) \(\displaystyle\frac{\pi}{8}\) — Incorrect. It takes t only from 0 to 1, which covers just \([0, \frac{\pi}{4}]\); for x up to \(\frac{\pi}{2}\), t runs from 1 to −1.
Option (4) \(\displaystyle\frac{1}{2}\) — Incorrect. The antiderivative is \(\tan^{-1}t\), so the value is a multiple of \(\pi\), not a plain number.