Mathematics

$\displaystyle \int{1\over {\sin x{{\cos }^3}x}}dx$

SOLUTION
Solution :-

$\displaystyle I= \int \frac{dx}{sinx\, cos^{3}x}$
$\displaystyle I= \int sin^{-1}x.cos^{-3}xdx$

Divide Num & Denominator by $cos^{4}x$
$\displaystyle \therefore I= \int \dfrac{1/cos^{4}x.dx}{\dfrac{sinx.cos^{3}x}{cos^{4}x}}$
$\displaystyle =\int \frac{sec^{4}xdx}{tan x}$

Let $tan x= t$

$\Rightarrow sec^{2}xdx= dt$
$\displaystyle \therefore I= \int \frac{sec^{2}x.dt}{t}$
$\displaystyle =\int \frac{(1+tan^{2}x).dt}{t}$
$\displaystyle =\int \frac{(1+t^{2})dt}{t}$

$\displaystyle =\int (\frac{1}{t}+t)dt$
$\displaystyle = logt+\frac{t^{2}}{2}+c$

$\displaystyle = log\left | tanx \right |+\frac{tan^{2}x}{2}+c$

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

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