Mathematics

# Solve : $\displaystyle\int sec^{-1} \sqrt{x} \, dx$

##### SOLUTION
$\displaystyle\int \sec^{-1}\sqrt{x}dx$
Let $x=\cos^2t$ here $t=\cos^{-1}\sqrt{x}$
$dx=-2\cos t\sin tdt$
Put there value
$-\displaystyle\int 2\sec^{-1}\sqrt{\cos^2x}\cos t\sin t dt$
$=-2\displaystyle\int \sec^{-1}\cos x\cos t\sin t dt$
$=-2\displaystyle\int x\cos t\sin tdt$
$=-2\displaystyle\int \cos^2t\cos t\sin tdt$
$=-2\displaystyle\int \cos^3t\sin tdt$
Let $\cos t=m$
$-\sin tdt=dm$
$=-2\displaystyle\int m^3dm$
$=-2\left[\dfrac{m^4}{4}\right]+c$
$=-\dfrac{\cos^4t}{2}+c=-\dfrac{\cos^4(\cos^{-1}\sqrt{x})}{2}+c$.

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 114

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