#### Need solution for RD Sharma Maths Class 12 Chapter Definite Integrals exercise 19.2 question 18.

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

Hint: We use indefinite integral formula then put limits to solve this integral.

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

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

Put $\sin2x=t$

$2\sin x \cos x=dt$

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

When x=0  then t=0  and when $x=\frac{\pi}{2}$  then t=1

\begin{aligned} &I=\int_{0}^{1} \frac{1}{1+t^{2}} \frac{d t}{2} \\ &=\frac{1}{2} \int_{0}^{1} \frac{1}{1+t^{2}} d t \\ &=\frac{1}{2}\left[\tan ^{-1}(t)\right]_{0}^{1} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}\right] \end{aligned}

\begin{aligned} &=\frac{1}{2}\left[\tan ^{-1} 1-\tan ^{-1} 0\right]\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\tan ^{-1} 1=\frac{\pi}{4}, \tan ^{-1} 0=0\right] \\ &=\frac{1}{2}\left[\frac{\pi}{4}-0\right] \\ &=\frac{1}{2} \cdot \frac{\pi}{4} \\ &=\frac{\pi}{8} \end{aligned}