Mathematics

# Find $\int{\dfrac{1}{\sin x\cos^{3}x}dx}$

##### SOLUTION
$\displaystyle\int \dfrac{1}{\sin x\cos^3x}dx$
$=\displaystyle\int \dfrac{1/\cos^4x}{\dfrac{\sin x\cos^3x}{\cos^4x}}dx$
$=\displaystyle\int\dfrac{(1+\tan^2x)\sec^2x}{\tan x}dx$
Now $\tan x=t\rightarrow \sec^3xdx=dt$
$\Rightarrow \displaystyle\int \dfrac{1+t^2}{t}=dt$
$\Rightarrow \displaystyle\int \dfrac{1}{t}i+dt$
$\Rightarrow log t+\dfrac{t^2}{2}+c$
$\Rightarrow log \tan t+\dfrac{\tan^2x}{2}+c$.

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

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