Mathematics

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

##### SOLUTION
Put $\cos^{-1}x = t$
$\Rightarrow x = \cos t$
$\Rightarrow dx = - \sin t dt$

$\therefore \displaystyle\int \dfrac{\cos^{-1}x}{x^2} dx = \int \dfrac{t}{\cos^2 t} (-\sin t) dt$

$= - \displaystyle\int \dfrac{t}{I} \dfrac{(\sec t.\tan t)dt}{II}$

$= - t. \sec t - \displaystyle\int 1. \sec t dt$

$= - t \sec t + \log |\sec t + \tan t| + c$

$= \dfrac{-t}{\cos t} + \log \left | \dfrac{1 + \sin t}{\cos t} \right | + c$

$= \dfrac{- \cos^{-1}x}{x} + \log \left | \dfrac{1 + \sin (\cos^{-1} x)}{x} \right | + c$

$= \dfrac{-\cos^{-1}x}{x} + \log \left | \dfrac{1 + \sqrt{1 - x^2}}{x} \right | + c$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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