First slide

Evaluation of definite integrals

Question

$\int_{0}^{\frac{π}{4}} (\sqrt{\tan x} + \sqrt{\cot x}) d x =$

Difficult

A

$π$
B

$\frac{π}{2}$
C

$\frac{π}{\sqrt{2}}$
D

$\frac{π}{2 \sqrt{2}}$

Solution

$I = \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{\sqrt{\sin x \cos x}} d x = \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{(\sin x \cos x) d x}{\sqrt{1 - {(\sin x - \cos x)}^{2}}}$
$= \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{d (\sin x - \cos x)}{\sqrt{1 - {(\sin x - \cos x)}^{2}}}$

$= \sqrt{2} \sin^{- 1} {(\sin x - \cos x)}_{0}^{\frac{π}{4}} = \frac{π}{\sqrt{2}}$

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