Mathematics

# Evaluate: $\displaystyle\int {\dfrac{1}{{\sin x{{\cos }^2}x}}dx}$

##### SOLUTION
$\displaystyle \int \dfrac{1}{\sin x \cos^2 x} dx$
$= \displaystyle\int \dfrac{\sin^2 x + \cos^2 x}{\sin x \cos^2 x} dx$     [using $\sin^2 x + \cos^2 x = 1$]
$= \displaystyle\int \dfrac{\sin x}{\cos^2 x} dx + \int \dfrac{1}{\sin x} dx$
for the first integral, put $\cos x = t$
$\Rightarrow - \sin x \, dx = dt$
$\Rightarrow dx = \dfrac{-1}{\sin x} dt$
$= \displaystyle\int \dfrac{-dt}{t^2} + \int cosec \, x \, dx$
$= \dfrac{-(t)^{-2 + 1}}{-2 + 1} + (-ln |cosec \, x + \cot x|)$
$= \dfrac{1}{t} - (ln |cosec x + cot x|)$
$= \dfrac{1}{\cos x} - ln |cosec x + cot x|$
$= \sec x - ln |cosec x + \cot x|$

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

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