Mathematics

# $\displaystyle\int \sec^{\dfrac{2}{3}}x cosec^{\dfrac{4}{3}}xdx$ is equal to?

$-3\cot^{\dfrac{1}{3}}x+c$

##### SOLUTION
$I=\displaystyle\int (\sec x)^{2/3}\cdot(cosec x)^{4/3}dx$
$=\displaystyle\int\dfrac{1}{(\sin x)^{4/3}\cdot (\cos x)^{2/3}}dx$
Multiplying numerator and denominator by $cosec^2x$, we get
$I=\displaystyle\int \dfrac{cosec^2x}{(cot x)^{2/3}}dx$
Let $\cot x=t^3$
$\Rightarrow cosec^2xdx=-3t^2dt$
Hence $I=-3\displaystyle\int \dfrac{t^2dt}{t^2}=-3t+C=-3(\cot x)^{1/3}+C$.

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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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