First slide

Introduction to integration

Question

$\int (\sqrt{\tan x} + \sqrt{\cot x}) d x$ is equal to

Difficult

A

$\sqrt{2} \sin^{- 1} (\sin x - \cos x) + c$
B

$\sqrt{2} \sin^{- 1} (\sin x + \cos x) + c$
C

$\sqrt{2} \tan^{- 1} (\sin x - \cos x) + c$
D

$\sqrt{2} \tan^{- 1} (\sin x + \cos x) + c$

Solution

$\int (\sqrt{\tan x} + \sqrt{\cot x}) d x$
$\int (\frac{\sqrt{\sin x}}{\sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}) d x$

$\int (\frac{\sin x + \cos x}{\sqrt{\sin x \cos x}}) d x$

$\int \frac{\sqrt{2} (\sin x + \cos x)}{\sqrt{2 \sin x \cos x}} d x$

$\sqrt{2} \int \frac{\sin x + \cos x}{\sqrt{1 - {(\sin x - \cos x)}^{2}}} d x$

Put $\sin x - \cos x = t$

$\Rightarrow (\cos x + \sin x) d x = d t$

$= \sqrt{2} \int \frac{d t}{\sqrt{1 - t^{2}}}$

$= \sqrt{2} \sin^{- 1} t + C$

$= \sqrt{2} \sin^{- 1} (\sin x - \cos x) + C$

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