MCQ
$\int_0^\pi \sin ^2 x d x$ is equal to
  • A
    $\pi$
  • $\frac{\pi}{2}$
  • C
    $0$
  • D
    $\frac{\pi}{3}$

Answer

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

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