#### Explain solution for  RD Sharma Class 12 Chapter 19 Definite Integrals Exercise 19.4 (b) Question 22 maths textbook solution.

Answer:- $\frac{\pi^{2}}{16}$

Hints:-  You must know the integration rules of trigonometric functions and its limits.

Given:-  $\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$

Solution : $\int_{0}^{\pi / 2} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$

\begin{aligned} &I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cdot \sin \left(\frac{\pi}{2}-x\right) \cdot \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x \\ &I=\int_{0}^{\pi / 2} \frac{\left(\frac{\pi}{2}-x\right) \cdot \cos x \cdot \sin x}{\cos ^{4} x+\sin ^{4} x} d x \end{aligned}

\begin{aligned} &I=\frac{\pi}{2} \int_{0}^{\pi / 2} \frac{\cos x \cdot \sin x}{\sin ^{4} x+\cos ^{4} x} d x-\int_{0}^{\pi / 2} \frac{x \cdot \cos x \cdot \sin x}{\sin ^{4} x+\cos ^{4} x} d x \\ &I=\frac{\pi}{2 \times 2} \int_{0}^{\pi / 2} \frac{2 \tan x \cdot \sec ^{2} x}{1+\left(\tan ^{2} x\right)^{2}} d x-I \end{aligned}

Let $\tan^{2}x = z$

\begin{aligned} &2 \tan x \sec ^{2} x \cdot d x=d z \\ &x=0, z=0 \\ &x=\frac{\pi}{2}, z=\infty \\ &2 I=\frac{\pi}{4} \int_{0}^{\infty} \frac{d z}{1+z^{2}} \end{aligned}

\begin{aligned} &I=\frac{\pi}{8} \int_{0}^{\infty}\left[\tan ^{-1} z\right]_{0}^{\infty} \\ &I=\frac{\pi}{8} \int_{0}^{\infty}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \\ &I=\frac{\pi}{8}\left[\frac{\pi}{2}-0\right] \\ &=\frac{\pi^{2}}{16} \end{aligned}