Question
By using the properties of definite integrals, evaluate the integral $\int\limits_{\frac{{ - \pi }}{2}}^{\frac{\pi }{2}} {{{\sin }^2}xdx}$
${\because \int\limits_{ - a}^a {f\left( x \right)dx = 2\int\limits_0^a {f\left( x \right)dx,} } }$ when f(x) is even function]
$\Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\sin }^2}\left( {\frac{\pi }{2} - x} \right)dx} $
$\left[ {\because \int\limits_0^a {f\left( x \right)dx = \int\limits_0^a {f\left( {a - x} \right)dx = } } } \right]$
$\Rightarrow I = 2\int\limits_0^{\frac{\pi }{2}} {{{\cos }^2}xdx} $ …(ii)
Adding eq. (i) and (ii),
$2I = 2\int\limits_0^{\frac{\pi }{2}} {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)dx} $
$= 2\int\limits_0^{\frac{\pi }{2}} {1dx} $
$= 2\left( x \right)_0^{\frac{\pi }{2}}$
$ = 2.\frac{\pi }{2} = \pi $
$\Rightarrow I = \frac{\pi }{2}$
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f2
Also, show that fof ≠ f2