Question
Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$
✓
Answer
Let $f(x) = \sin^2x$
$f(-x) = \sin^2(-x) = \sin^2x = f(x)$
$\therefore$ function is even
$\therefore \int_{ - \pi /4}^{\pi /4} {{{\sin }^2}xdx = 2\int_0^{\pi /4} {{{\sin }^2}xdx} }$
$= \int_\limits0^{\pi /4} {2\left( {\frac{{1 - \cos 2x}}{2}} \right)dx} $
$= \int_\limits0^{\pi /4} {\left( {{{1 - \cos 2x}}{}} \right)dx} $
$= \left[ {x - \frac{{\sin 2x}}{2}} \right]_0^{\pi /4}$
$ = \frac{\pi }{4} - \frac{1}{2}$
$f(-x) = \sin^2(-x) = \sin^2x = f(x)$
$\therefore$ function is even
$\therefore \int_{ - \pi /4}^{\pi /4} {{{\sin }^2}xdx = 2\int_0^{\pi /4} {{{\sin }^2}xdx} }$
$= \int_\limits0^{\pi /4} {2\left( {\frac{{1 - \cos 2x}}{2}} \right)dx} $
$= \int_\limits0^{\pi /4} {\left( {{{1 - \cos 2x}}{}} \right)dx} $
$= \left[ {x - \frac{{\sin 2x}}{2}} \right]_0^{\pi /4}$
$ = \frac{\pi }{4} - \frac{1}{2}$
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