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Question:

$\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$

Solution:

$I=\int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$     ...(1)

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \cos ^{2}\left(\frac{\pi}{2}-x\right) d x$    $\left(\int_{0}^{0} f(x) d x=\int_{0}^{0} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \sin ^{2} x d x$   ...(2)

Adding (1) and (2), we obtain

$2 I=\int_{2}^{\frac{\pi}{2}}\left(\sin ^{2} x+\cos ^{2} x\right) d x$

$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 1 d x$

$\Rightarrow 2 I=[x]_{0}^{\frac{\pi}{2}}$

$\Rightarrow 2 I=\frac{\pi}{2}$

$\Rightarrow I=\frac{\pi}{4}$

 

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