MCQ
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x=$
  • A
    $\frac{\pi}{3}$
  • B
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{6}$
  • $\frac{\pi}{2}$

Answer

Correct option: D.
$\frac{\pi}{2}$
(D)
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x=2 \int_0^{\frac{\pi}{2}} \sin ^2 x d x$ $\ldots \left[\because \sin ^2 x\right.$ is an even function $]$
If n is a positive integer, then
$\int_0^{\pi / 2} \sin ^{n} x d x=\int_0^{\pi / 2} \cos ^{n} x d x$
$=\left\{\begin{array}{l}\frac{(n-1)(n-3) \ldots .2}{n(n-2) \ldots .3}, \text { when } n \text { is odd } \\ \frac{(n-1)(n-3) \ldots .1}{n(n-2) \ldots .2} \cdot \frac{\pi}{2}, \text { when } n \text { is even }\end{array}\right.$
$\therefore \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^2 x d x=2\left(\frac{1}{2} \cdot \frac{\pi}{2}\right)=\frac{\pi}{2} $

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