Mathematics

# The value of the integral $\int_{0}^{\pi/2}\dfrac{3\sqrt{cos \theta}}{(\sqrt{\cos \theta} +\sqrt{\sin \theta })^5}d\theta$equal

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##### SOLUTION
$I=3\displaystyle \int_0^{\pi/2}\dfrac{\sqrt{cos \theta }}{(\sqrt{\sin \theta }+\sqrt{\cos \theta})^5}=3 \int_0^{\pi/2}\dfrac{3\sqrt{\sin \theta }d\theta }{(\sqrt{\cos \theta}+\sqrt{\sin \theta })^5}$
$\Rightarrow \displaystyle \int_0^{\pi/2}\dfrac{3\sqrt{cos \theta }}{(\sqrt{\sin \theta }+\sqrt{\cos \theta})^5}=3 \int_0^{\pi/2}\dfrac{\sqrt{\sin \theta }d\theta }{(\sqrt{\cos \theta}+\sqrt{\sin \theta })^5}$
$\Rightarrow 2I=3\int_0^{\pi/2}\dfrac{d\theta }{(\sqrt{\sin \theta }+\sqrt{\cos \theta})^4}$
$\dfrac{2I}{3}=\int_0^{\pi/2}\dfrac{\sec^2 \theta d \theta }{(\sqrt{\tan \theta }+1)^4}$
Let $\tan \theta =t^2\Rightarrow \dfrac{I}{3}=\displaystyle \int_0^{infty}\left[\dfrac{1}{(t+1)^3}-\dfrac{1}{(t+1)^4}\right]dt$
$I=\left|\dfrac{-3}{2(t+1)^2}+\dfrac{1}{(t+1)^3}\right|_0^{\infty}=\dfrac{3}{2}-1=\dfrac{1}{2}$

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