MCQ
The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} d x$ is
  • A
    $\frac{\pi}{2}$
  • B
    $4 \pi$
  • $\frac{\pi}{4}$
  • D
    $\frac{\pi}{8}$

Answer

Correct option: C.
$\frac{\pi}{4}$
(C)
$\text { Let } I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^x} d x$ ...(i)
$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sin ^2 x}{1+2^{-x}} d x$ ....(ii)
$\ldots\left[\because \int_{ a }^{ b } f (x) d x=\int_{ a }^{ b } f ( a + b -x) d x\right]$
Adding (i) and (ii), we get
$2 I =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x$
$\Rightarrow 2 I =2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x$
...$\left[\begin{array}{l}\because \int_{- a }^{ a } f (x) d x=2 \int_0^{ a } f (x) d x \\ \text { if } f (x) \text { is an even function }\end{array}\right]$
$\Rightarrow I =\int_0^{\frac{\pi}{2}}\left(\frac{1-\cos 2 x}{2}\right) d x$
$=\frac{1}{2}\left[x-\frac{\sin 2 x}{2}\right]_0^{\frac{\pi}{2}}=\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}$

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