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
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