Mathematics

$\int {\dfrac {\cos 2x}{\sin x}}dx$

SOLUTION
$I=\int {\dfrac {\cos 2x}{\sin x}}dx\\\because \cos 2x=\cos^2 x-\sin^2 x\\=1-\sin^2x-\sin^2x\\=1-2\sin^2x\\\therefore I=\int{\dfrac {1-2\sin^2x}{\sin x}}dx=\int {\left[\dfrac 1{\sin x}-\dfrac {2\sin^2x}{\sin x}\right]dx}\\\implies=\int {\csc x}dx-2\int {\sin x}dx\\\implies=-\ln{|\csc x+\cot x|}-2(-\cos x)+C\\I=2\cos x-\ln{|\csc x+\cot x|}+C$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 111

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