Question

The value of ${\int }_{-1}^1\left(\frac{x^2+\sin x}{1+x^2}\right)dx$, is equal to

• A. $2-\pi$
• B. $\pi -2$
• C. $2-\frac{\pi }{2}$
• D. none of these

Answer 1

Answer: (C)

$2-\frac{\pi }{2}$

Let $I={\int }_{-1}^1\left(\frac{x^2+\sin x}{1+x^2}\right)dx={\int }_{-1}^1(\frac{x^2}{1+x^2}+\frac{\sin x}{1+x^2})dx$
Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_{-1}^1(\frac{x^2}{1+x^2}-\frac{\sin x}{1+x^2})dx$
$\therefore 2I={\int }_{-1}^1\left(\frac{x^2}{1+x^2}\right)dx=2{\int }_{-1}^1(1-\frac{1}{1+x^2})dx$
$I={[x-{\tan }^{-1}x]}_{-1}^1=2-\frac{\pi }{2}$