Mathematics

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


SOLUTION

Consider the given integral.

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

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

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

Therefore,

$$ I=\int{t\sin t}dt $$

$$ I=t\left( -\cos t \right)-\int{1\left( -\cos t \right)dt} $$

$$ I=-t\cos t+\int{\cos tdt} $$

$$ I=-t\cos t+\sin t+C $$

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

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

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

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

Hence, this is the answer.

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Subjective Medium Published on 17th 09, 2020
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