Discussion Forum

The value  of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2}x}{1+2^{x}}$ is

Answer:

Explanation:

Key Idea 

 Use property = $\int_{a}^{b} f( x) dx=\int_{a}^{b} f( a+b-x)dx$

   Let $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^{2}x}{1+2^{x}} dx$

$\Rightarrow$   I $= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin^{2}(\frac{-\pi}{2}+\frac{\pi}{2}-x)}{1+2^{\frac{-\pi}{2}+\frac{\pi}{2}-x}} dx$

                                               [ $\because$  $\int_{a}^{b} f( x) dx=\int_{a}^{b} f( a+b-x)dx$ ]

$\Rightarrow$      $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin^{2}x}{1+2^{-x}}dx$

$\Rightarrow$   $I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{2^{x}\sin^{2}x}{2^{x}+1^{}}dx$

$\Rightarrow$    $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}x(\frac{2^{x}+1}{2^{x}+1})dx$

$\Rightarrow$     $2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^{2}x dx$

$\Rightarrow$   $2I=2\int_{0}^{\frac{\pi}{2}}\sin^{2}x dx$       [ $\because\sin^{2}x$ is an even function ]

$\Rightarrow$  $I= \int_{0}^{\frac{\pi}{2}} \sin^{2}x dx$

$\Rightarrow$  $I= \int_{0}^{\frac{\pi}{2}} \cos^{2}x dx$

                                              [ $\because \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a-x)dx$]

$\Rightarrow$   2I=$\int_{0}^{\frac{\pi}{2}} dx$

$\Rightarrow$  $2I= [ x] _{0}^{\frac{\pi}{2}}=I=\frac{\pi}{4}$