Question

The value of $\stackrel{\pi /2}{\underset{-\pi /2}{\int }}\frac{x^2}{1+\tan x+\sqrt{1+{\tan }^{2}x}}dx$ is

• A. ${\pi }^3$
• B. $\frac{{\pi }^3}{12}$
• C. $\frac{{\pi }^3}{24}$
• D. $\frac{{\pi }^3}{48}$

Answer 1

The correct option is C $\frac{{\pi }^3}{24}$

$I=\stackrel{\pi /2}{\underset{-\pi /2}{\int }}\frac{x^2}{1+\tan x+\sqrt{1+{\tan }^{2}x}}dx$
Using the property,
$\stackrel{b}{\underset{a}{\int }}f\left(x\right)dx=\stackrel{b}{\underset{a}{\int }}f(a+b-x)dx$
$I=\stackrel{\pi /2}{\underset{-\pi /2}{\int }}\frac{x^2}{1-\tan x+\sqrt{1+{\tan }^{2}x}}dx\text{On adding these two integrals we get,}2I=\stackrel{\pi /2}{\underset{-\pi /2}{\int }}\frac{x^2(2+2\sqrt{1+{\tan }^{2}x})}{(1+\sqrt{1+{\tan }^{2}x}{)}^2-{\tan }^{2}x}dx2I=\stackrel{\pi /2}{\underset{-\pi /2}{\int }}{x}^{2}dx=[\frac{x^3}{3}{]}_{-\pi /2}^{\pi /2}2I=\frac{1}{3}\times \frac{2{\pi }^3}{8}\Rightarrow I=\frac{{\pi }^3}{24}$