Definite Integration Question 82
Question 82
- Evaluate $\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$.
$\left(1985,2 \frac{1}{2} \mathrm{M}\right)$
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Answer:
Correct Answer: 84. $\frac{\pi^{2}}{16}$
Solution:
- Let $I=\int_{0}^{\pi / 2} \frac{x \sin x \cdot \cos x}{\cos ^{4} x+\sin ^{4} x} d x$
$\Rightarrow I=\int_{0}^{\pi / 2} \frac{\frac{\pi}{2}-x \sin \frac{\pi}{2}-x \cdot \cos \frac{\pi}{2}-x}{\sin ^{4} \frac{\pi}{2}-x+\cos ^{4} \frac{\pi}{2}-x} d x$
$\Rightarrow I=\int_{0}^{\pi / 2} \frac{\frac{\pi}{2}-x \cdot \sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$
$\Rightarrow I=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x-\int_{0}^{\pi / 2} \frac{x \sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x$
$=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\sin x \cdot \cos x}{\sin ^{4} x+\cos ^{4} x} d x-I$
$\Rightarrow 2 I=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\tan x \cdot \sec ^{2} x}{\tan ^{4} x+1} d x$
$\Rightarrow 2 I=\frac{\pi}{2} \cdot \frac{1}{2} \int_{0}^{\pi / 2} \frac{1}{1+\left(\tan ^{2} x\right)^{2}} d\left(\tan ^{2} x\right)$
$\Rightarrow 2 I=\frac{\pi}{4} \cdot\left[\tan ^{-1} t\right]_{0}^{\infty}=\frac{\pi}{4}\left(\tan ^{-1} \infty-\tan ^{-1} 0\right)$
$\Rightarrow \quad I=\frac{\pi^{2}}{16}$
[where, $t=\tan ^{2} x$ ]
