Evaluate _ - 4 ^ 4 ^2 x d x. - Vidyadip

Question

Evaluate $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x$.

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Answer

Here $\sin ^2 x$ is an even function.
So, $\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}} \sin ^2 x d x=2 \int_0^{\frac{\pi}{4}} \sin ^2 x d x$
$
\left(\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x\right)
$
$\begin{array}{l}=2 \int_0^{\frac{\pi}{4}} \frac{1-\cos 2 x}{2} d x \\ =2 \int_0^{\frac{\pi}{4}}(1-\cos 2 x) d x \\ =\left(x-\frac{1}{2} \sin 2 x\right)_0^{\frac{\pi}{4}} \\ =\left(\frac{\pi}{4}-\frac{1}{2} \sin \frac{\pi}{2}\right)-0=\frac{\pi}{4}-\frac{1}{2} \text { }\end{array}$

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