First slide

Evaluation of definite integrals

Question

The value of the integral $\int_{0}^{π / 2} \frac{3 \sqrt{\cos θ}}{{(\sqrt{\cos θ} + \sqrt{\sin θ})}^{5}} d θ$ is equal to

Moderate

A

$\frac{1}{2}$
B

$1$
C

$\frac{3}{2}$
D

$2$

Solution

$\begin{array}{l} I = \int_{0}^{π / 2} \frac{3 \sqrt{\cos θ}}{(\sqrt{\cos θ} + \sqrt{\sin θ})^{5}} d θ = \int_{0}^{π / 2} \frac{3 \sqrt{\sin θ}}{(\sqrt{\sin θ} + \sqrt{\cos θ})^{5}} d θ \\ 2 I = \int_{0}^{π / 2} \frac{3}{(\sqrt{\sin θ} + \sqrt{\cos θ})^{4}} d θ = \int_{0}^{π / 2} \frac{3 \sec^{2} θ}{(1 + \sqrt{Tan θ})^{4}} d θ \\ Put, Tan θ = t^{2} \Rightarrow \sec^{2} θ d θ \Rightarrow 2 t d t \\ = \int_{0}^{\infty} \frac{6 t}{(1 + t)^{4}} d t = 6 \int_{0}^{\infty} (\frac{1}{(1 + t)^{3}} - \frac{1}{(1 + t)^{4}}) d t \\ \Rightarrow I = 3 {(\frac{(1 + t)^{- 3}}{3} - \frac{(1 + t)^{- 2}}{2})}_{0}^{\infty} \Rightarrow I = 3 (\frac{1}{3} (0 - 1) - \frac{1}{2} (0 - 1)) = \frac{1}{2} \end{array}$

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