Question

The value of ${\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}\frac{{\sin }^{2}x}{1+2^x}dx$ is $:$

• A. $4\pi$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{8}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

Using,

$I={\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a-b-x)dx$

$I_2={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2x}dx$

Remember $I_1={\int }_{-a}^{a}f\left(x\right)dx$

$\Rightarrow I_1={\int }_0^a\left[f\right(x)+f(-x\left)\right]dx$

$I_2={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}x}{1+2^x}dx={\int }_0^{\pi /2}(\frac{{\sin }^{2}x}{1+2^x}+\frac{{\sin }^2(-x)}{1+2^{-x}})dx$

$I_2={\int }_0^{\pi /2}{\sin }^{2}x\left(\frac{1+2^x}{1+2^x}\right)dx$

$I_2={\int }_0^{\pi /2}{\sin }^{2}xdx$ .....(1)

$I_2={\int }_0^{\pi /2}{\sin }^2(\frac{\pi }{2}-x)dx={\int }_0^{\pi /2}{\cos }^{2}xdx$ .....(2)

(1) + (2)

$I_2={\int }_0^{\pi /2}1.dx$

$I_2=\frac{\pi /2}{2}$

$I_2=\frac{\pi }{4}$