Mathematics

# $\int \dfrac{1}{sinxcos^{2}x}=dx$.

$secx+log\mid cosecx-cot x\mid +c$

##### SOLUTION
$\displaystyle\int \dfrac{1}{\sin x\cos^2x}dx=\displaystyle\int \dfrac{\sin^2x+\cos^2x}{\sin x\cos^2x}dx$
$=\displaystyle\int \dfrac{\sin^2x}{\sin x\cos^2x}dx+\displaystyle\int \dfrac{\cos^2x}{\sin x\cos^2x}dx$
$=\displaystyle\int \dfrac{\sin x}{\cos^2x}dx+\displaystyle\int \dfrac{1}{\sin x}dx$
$=\displaystyle\int \sec x\cdot \tan x dx+\displaystyle\int cosec xdx$
$=\sec x+ln|cosec x-\cot x|+c$.

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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