#### Please solve RD Sharma class 12 chapter 19 Definite Integrals exercise Fill in the blanks question 21 maths textbook solution

Answer: $\frac{\pi }{4}$

Hint: Using $\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$

Given: $\int_{0}^{\pi / 2} \frac{1}{1+\tan ^{3} x} d x$

Solution:

\begin{aligned} &\mathrm{I}=\int_{0}^{\pi / 2} \frac{1}{1+\tan ^{3} x} \mathrm{dx} \end{aligned}                                    .........(1)

$\Rightarrow I=\int_{0}^{\pi / 2} \frac{1}{1+\tan ^{3}\left(\frac{\pi}{2}-x\right)} d x$

$\tan \left(\frac{\pi}{2}-x\right)=\cot x=\frac{1}{\tan x}$

$\therefore I=\int_{0}^{\pi / 2} \frac{1}{1+\frac{1}{\tan ^{3} x}}=\int_{0}^{\pi / 2} \frac{\tan ^{3} x}{\tan ^{3} x+1}$...........(2)

\begin{aligned} &2 I=\int_{0}^{\pi / 2} \frac{1+\tan ^{3} x}{1+\tan ^{3} x} d x \\\\ &\quad=\int_{0}^{\pi / 2} d x \end{aligned}

\begin{aligned} &=[x]_{0}^{\pi / 2} \\\\ &=\frac{\pi}{2}-0 \end{aligned}

\begin{aligned} &2 \mathrm{I}=\frac{\pi}{2} \\\\ &\mathrm{I}=\frac{\pi}{4} \end{aligned}