Mathematics

Integrate:
$$\displaystyle \int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$$


SOLUTION

Consider the given integral.

$$I=\int{\dfrac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}}dx$$

 

Let $$t={{\sin }^{-1}}x$$

$$dt=\dfrac{dx}{\sqrt{1-{{x}^{2}}}}$$

 

Therefore,

$$ I=\int{tdt} $$

$$ I=\dfrac{{{t}^{2}}}{2}+C $$

 

On putting the value of $$t$$, we get

$$I=\dfrac{{{\left( {{\sin }^{-1}}x \right)}^{2}}}{2}+C$$

 

Hence, this is the answer.

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