The value of the integral int 0 π / 2 frac 3 √(cos θ) (√(cos θ)+√(sin θ)) 5 d θ equals \ \ \ \

Mathematics · Definite Integration

JEE Advanced 2019 — Paper 2 — Question 28

The value of the integral

\int_{0}^{\pi / 2} \frac{3 \sqrt{\cos \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^{5}} \mathrm{~d} \theta \text { equals }

$\_\_\_\_$

Numerical answer — enter this value.

\begin{aligned} & \mathrm{I}=\int_{0}^{\pi / 2} \frac{3 \sqrt{\cos \theta} \mathrm{~d} \theta}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^{5}} \\& \quad=\int_{0}^{\pi / 2} \frac{3 \sqrt{\sin \theta}}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^{5}} d \theta \\& \Rightarrow \quad 2 \mathrm{I}=3 \int_{0}^{\pi / 2} \frac{d \theta}{(\sqrt{\cos \theta}+\sqrt{\sin \theta})^{4}}=\int_{0}^{\pi / 2} \frac{3 d \theta}{\cos ^{2} \theta(1+\sqrt{\tan \theta})^{4}} \\& \Rightarrow \quad \mathrm{I}=\frac{3}{2} \int_{0}^{\pi / 2} \frac{\sec ^{2} \theta \mathrm{~d} \theta}{(1+\sqrt{\tan \theta})^{4}} \end{aligned}

\text { Let } 1+\sqrt{\tan \theta}=\mathrm{t} \Rightarrow \frac{1}{2 \sqrt{\tan \theta}} \sec ^{2} \theta \mathrm{~d} \theta=\mathrm{dt}

\begin{aligned} & \mathrm{I}=\frac{3}{2} \int_{1}^{\infty} \frac{2(\mathrm{t}-1)}{\mathrm{t}^{4}} \mathrm{dt}=\frac{3}{2}\left|\frac{2}{-2 \mathrm{t}^{2}}+\frac{2}{3 \mathrm{t}^{3}}\right|_{1}^{\infty} \\& =\frac{3}{2} \theta \frac{2}{2}-\frac{2}{3} \theta=\frac{1}{2}=0.5 \end{aligned}

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