Question

The value of definite integral $\underset{-\pi }{\overset{\pi }{\int }}\frac{x^2}{1+\sin x+\sqrt{1+{\sin }^{2}x}}dx$, is

• A. $\frac{3{\pi }^3}{2}$
• B. $\frac{{\pi }^3}{3}$
• C. $\frac{2{\pi }^3}{3}$
• D. $\frac{{\pi }^3}{6}$

Answer 1

Answer: (B)

$\therefore 2I=\underset{-\pi }{\overset{\pi }{\int }}\frac{x^2\left(2+2\sqrt{1+{\sin }^{2}x}\right)}{{\left(1+\sqrt{1+{\sin }^{2}x}\right)}^2-{\sin }^{2}x}dx$
So, $I=\underset{-\pi }{\overset{\pi }{\int }}\frac{x^2\left(1+\sqrt{1+{\sin }^{2}x}\right)}{2\left(1+\sqrt{1+{\sin }^{2}x}\right)}dx=\frac{1}{2}\underset{-\pi }{\overset{\pi }{\int }}{x}^{2}dx=\frac{{\pi }^3}{3}$.