Question

Evaluate : $\int \frac{{\cos }^{-1}x}{x^2}dx$

Answer 1

Put ${\cos }^{-1}x=t$
$\Rightarrow x=\cos t$
$\Rightarrow dx=-\sin tdt$

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

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

$=-t.\sec t-\int 1.\sec tdt$

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

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

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

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