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Question:

$\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x}$

Solution:

Let $I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$    …(1)

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5}\left(\frac{\pi}{2}-x\right)}{\sin ^{5}\left(\frac{\pi}{2}-x\right)+\cos ^{5}\left(\frac{\pi}{2}-x\right)} d x$        $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$

$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$   …(2)

Adding (1) and (2), we obtain

$2 I=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{5} x+\cos ^{5} x}{\sin ^{5} x+\cos ^{5} x} d x$

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

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

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

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

 

 

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