#### Explain solution RD Sharma Class12 Chapter Definite Integrals exercise 19.2 question 26 maths.

Answer: $\frac{1}{8}$

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

Given: $\int_{0}^{\frac{\pi}{4}} \frac{\tan^3x}{1+\cos ^22x}dx$

Solution: $I=\int_{0}^{\frac{\pi}{4}} \frac{\tan^3x}{1+\cos ^22x}dx$

$=\int_{0}^{\frac{\pi}{4}} \frac{\tan^3x}{2\cos ^2x}dx$                                                    $\left [ 1+\cos 2\theta =2 \cos^2 \theta \right ]$

\begin{aligned} &=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \frac{\tan ^{3} x}{\cos ^{2} x} d x \\ &I=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} \tan ^{3} x \sec ^{2} x d x \end{aligned}

Put $\tan x=t$

$\sec^2 xdx=dt$

When x=0  then t=0  and when $x=\frac{\pi}{4}$  then t=1

\begin{aligned} I &=\frac{1}{2} \int_{0}^{1} t^{3} d t \\ &=\frac{1}{2}\left[\frac{t^{3+1}}{3+1}\right]_{0}^{1} \\ &=\frac{1}{2} \cdot \frac{1}{4}\left[t^{4}\right]_{0}^{1} \\ &=\frac{1}{8}\left[1^{4}-0^{4}\right] \\ &=\frac{1}{8}[1] \\ &=\frac{1}{8} \end{aligned}