#### Need solution for RD Sharma Maths Class 12 Chapter 19 Definite Integrals Excercise Revision Exercise Question 47

Answer:  $\frac{\pi^{2}}{16}$

Hint: To solve this we use  $a^{4}+b^{4}, \int_{0}^{a} f(x)$   form

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

Solution:

$I= \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x$         ………. (1)

\begin{aligned} &\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x \\ & \end{aligned}

$I=\int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right) \cos \left(\frac{\pi}{2}-x\right)}{\sin ^{4}\left(\frac{\pi}{2}-x\right)+\cos ^{4}\left(\frac{\pi}{2}-x\right)} d x$

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

$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{\cos ^{4} x+\sin ^{4} x} d x \\$

${\left[a^{4}+b^{4}=\left(a^{2}+b^{2}\right)^{2}-2 a^{2} b^{2}\right]}$

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

$I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\cos x \sin x}{1-(2 \cos x \cdot \sin x)^{2}} d x$

\begin{aligned} &\sin 2 x=2 \sin x \cdot \cos x \\ & \end{aligned}

$\sin x \cos x=\frac{\sin 2 x}{2}$

\begin{aligned} &I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin \frac{2 x}{2}}{1-\left(\frac{\sin 2 x}{2}\right)^{2}} d x \\ & \end{aligned}

$I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin \frac{2 x}{2}}{\left(\frac{2-\sin ^{2} 2 x}{2}\right)} d x$

\begin{aligned} &I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+1-\sin ^{2} x} d x \\ & \end{aligned}

$I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^{2} x} d x$

\begin{aligned} &\cos 2 x=t \\ & \end{aligned}

$-2 \sin 2 x=\frac{d t}{d x} \\$

$\sin 2 x d x=\frac{-d t}{2}$

\begin{aligned} &I=\frac{\pi}{4} \times \frac{-1}{2} \int \frac{d t}{1+t^{2}} \\ & \end{aligned}                         $\left[\begin{array}{l} x=0, t=1 \\ x=\frac{\pi}{2}, t=-1 \end{array}\right]$

$I=\frac{-\pi}{8} \int_{+1}^{-1} \frac{d t}{1+t^{2}}$

$\int_{a}^{b} f(x) d x=-\int_{b}^{a} f(x) d x \\$

$I=\frac{\pi}{8} \int_{-1}^{1} \frac{d t}{1+t^{2}}$

\begin{aligned} &I=\frac{\pi}{8}\left[\tan ^{-1} t\right]_{-1}^{1} \\ \end{aligned}

$I=\frac{\pi}{8}\left[\tan ^{-1} 1-\tan ^{-1}(-1)\right] \\$

$I=\frac{\pi}{8}\left[\frac{\pi}{4}+\frac{\pi}{4}\right] \\$

$I=\frac{\pi}{8}\left(\frac{2 \pi}{4}\right) \\$

$I=\frac{\pi^{2}}{16}$