Evaluate the definite integral _0^ ( ^2 x 2 - ^2 x 2 )dx - Vidyadip

Question

Evaluate the definite integral $\int\limits_0^\pi {\left( {{{\sin }^2}\frac{x}{2} - {{\cos }^2}\frac{x}{2}} \right)dx} $

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Answer

$\int\limits_0^\pi {\left( {{{\sin }^2}\frac{x}{2} - {{\cos }^2}\frac{x}{2}} \right)dx} $ $= \int\limits_0^\pi {\left( {\frac{{1 - \cos x}}{2} - \frac{{1 + \cos x}}{2}} \right)dx} $
$= \int\limits_0^\pi {\left( {\frac{{1 - \cos x - 1 - \cos x}}{2}} \right)dx} $
$ = \int\limits_0^\pi {\left( {\frac{{ - 2\cos x}}{2}} \right)dx} $
$ = - \int\limits_0^\pi {\cos xdx} $
$ = - \left( {\sin x} \right)_0^\pi $
$= - \left( {\sin \pi - \sin {0^o}} \right)$
= -(0 - 0) = 0

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