#### Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 11

Answer: $\frac{\pi }{8}$

Hint: Use $\int \frac{1}{1+x^{2}} d x$

Given: $\int_{0}^{\pi / 2} \frac{\sin x \cos x}{1+\sin ^{4} x} d x$

Solution:

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

Put

\begin{aligned} &\sin ^{2} x=t \\\\ &2 \sin x \cos x \; d x=d t \\\\ &\sin x \cos x \; d x=\frac{d t}{2} \end{aligned}

When $\mathrm{x}=0, \mathrm{t}=0$

When $\mathrm{x}=\frac{\pi}{2}, \mathrm{t}=1$

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

\begin{aligned} &=\int_{0}^{1} \frac{1}{1+t^{2}} \cdot \frac{d t}{2} \\\\ &=\frac{1}{2}\left[\tan ^{-1} t\right]_{0}^{1} \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \\\\ &=\frac{1}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(0)\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1}\left(\tan \frac{\pi}{4}\right)-\tan ^{-1}(\tan 0)\right] \\\\ &=\frac{1}{2}\left(\frac{\pi}{4}\right)-\frac{1}{2}(0) \\\\ &=\frac{\pi}{8} \end{aligned}