Mathematics

Evaluate:
$$ \int_{}^{} {\frac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx} $$


SOLUTION
Put x = sint   $$\Rightarrow \frac{{dx}}{{dt}} = \cos tdt$$

$$\therefore I = \int_{}^{} {\frac{{ - \cos tdt}}{{\sqrt {1 - {{\sin }^2}t} }} = \int_{}^{} {\frac{{ - \cos tdt}}{{{{\cos }^2}t}}} } $$

$$ =  - t = {\sin ^{ - 1}}x$$

$${\cos ^{ - 1}}x + {\sin ^{ - 1}}x = \frac{\pi }{2}$$

$$ \Rightarrow {\cos ^{ - 1}} - \frac{\pi }{2} =  - {\sin ^{ - 1}}x$$

$$\therefore t = {\cos ^{ - 1}}x - \frac{\pi }{2}$$

Here, $$C =  - \frac{\pi }{2}$$
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Subjective Easy Published on 17th 09, 2020
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