Value of \( \int_{0}^{\pi/2}\displaystyle\frac{\sin x\cos x}{\cos^4x+\sin^4x}\,dx \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{\pi}{4} \)
Let \( t=\sin^2x \), so \( dt=2\sin x\cos x\,dx \) and \( \cos^2x=1-t \). The limits change: x = 0 and x = π/2 become t = 0 and t = 1.
\( \displaystyle \cos^4x+\sin^4x=(1-t)^2+t^2 \)
\( \displaystyle =2t^2-2t+1 \)
\( \displaystyle =2\left[\left(t-\frac{1}{2}\right)^2+\frac{1}{4}\right] \)
\( \displaystyle I=\frac{1}{2}\int_{0}^{1}\frac{dt}{2\left[\left(t-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2\right]} \)
\( \displaystyle =\frac{1}{4}\int_{0}^{1}\frac{dt}{\left(t-\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2} \)
By the standard integral \( \int\frac{dt}{u^2+a^2}=\frac{1}{a}\tan^{-1}\frac{u}{a} \) with \( a=\frac{1}{2} \):
\( \displaystyle I=\frac{1}{4}\cdot 2\Big[\tan^{-1}(2t-1)\Big]_{0}^{1} \)
\( \displaystyle =\frac{1}{2}\left[\tan^{-1}1-\tan^{-1}(-1)\right] \)
\( \displaystyle =\frac{1}{2}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] \)
\( \displaystyle =\frac{\pi}{4} \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{\pi^2}{16} \) — Incorrect. \( \frac{\pi^2}{16} \) is the value of the well-known integral \( \int_{0}^{\pi/2}\frac{x\sin x\cos x}{\sin^4x+\cos^4x}\,dx \), which has an extra factor x in the numerator; this integral has none.
Option (2) \( \displaystyle\frac{\pi}{8} \) — Incorrect. The value is \( \frac{\pi}{4} \), twice \( \frac{\pi}{8} \).
Option (4) \( \displaystyle\frac{\pi^2}{8} \) — Incorrect. With no x in the numerator, the value cannot contain \( \pi^2 \); it is \( \frac{\pi}{4} \).