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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

$I= \int \limits_0^{\frac{\pi}{2}} \sin ^2 x dx $ find the value of I

\[\begin {array} {1 1} (a)\;\frac{\pi}{2} \\ (b)\;\frac{\pi}{4} \\ (c)\;1 \\ (d)\;0 \end {array}\]

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1 Answer

$I=\int \limits _0 ^ {\frac{\pi}{2}}$$ \bigg(\large\frac{1- \cos 2x}{2}\bigg)$$dx$
$I= \large\frac{1}{2} \bigg(\frac{\pi}{2} - \frac{\sin 2x}{2} \bigg)_0^{\pi/2}$
$\qquad= \large\frac{1}{2} \bigg(\large\frac{\pi}{2}-0\bigg)$
$\qquad=\large\frac{\pi}{4}$
Hence b is the correct answer.
answered Dec 20, 2013 by meena.p
 
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