Mathematics

# Find :  $\displaystyle\int \dfrac{sin x - cos x}{\sqrt{1 + sin \, 2x}} dx , \, 0 < x < \dfrac{\pi}{2}$

##### SOLUTION
$\sqrt{1+\sin 2{x}}=\sqrt{\sin^2 x+\cos ^2 x+2\sin x\cos x}\sqrt{(\sin x+\cos x)^2}=|\sin x+\cos x|$
$0<x<\dfrac{\pi}{2},\implies \sqrt{1+\sin 2{x}}=\sin x+\cos x$
$\displaystyle\int \dfrac{\sin x-\cos x}{\sqrt{1+\sin 2x}}d x=\int \dfrac{\sin x-\cos x}{\sin x+\cos x}d x=\int \dfrac{\tan x-1}{1+\tan x}d x=\int \tan \bigg(x-\dfrac{\pi}{4}\bigg)d x=\log \bigg|\text{sec}\bigg(x-\dfrac{\pi}{4}\bigg)\bigg|+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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