Mathematics

Integrate the following function:
$$\displaystyle \int { \dfrac { { d }^{ 2 } }{ { dx }^{ 2 } }  } \left( \sin ^{ -1 }{ x }  \right) dx$$


ANSWER

$$(1-x^2)^{\frac{-1}{2}}+c$$


SOLUTION
Consider, $$\displaystyle I= \int { \cfrac { { d }^{ 2 } }{ d{ x }^{ 2 } } \left( \sin ^{ -1 }{ x }  \right) dx } $$

$$\displaystyle I= \int { \cfrac { d }{ dx } \left( \cfrac { d }{ dx } \left( \sin ^{ -1 }{ x }  \right)  \right) dx } $$

$$\displaystyle I= \int { \cfrac { d }{ dx } \left( \cfrac { 1 }{ \sqrt { 1-{ x }^{ 2 } }  }  \right) dx } $$

$$\displaystyle I=\int { \cfrac { 1\left( \cfrac { 1 }{ 2\sqrt { 1-{ x }^{ 2 } }  }  \right) 2x }{ 1-{ x }^{ 2 } } dx } $$

$$\displaystyle I=\int { \cfrac { x }{ { \left( 1-{ x }^{ 2 } \right)  }^{ 3/2 } } dx } $$

$$\displaystyle \because 1-{ x }^{ 2 }=t;\quad -2xdx=dt$$

$$\displaystyle I=\int { \cfrac { -dt }{ 2{ t }^{ 3/2 } }  } =\cfrac {- 1 }{ 2 } \int { { t }^{ -3/2 }dt } $$

$$\displaystyle I=\cfrac { -1 }{ 2 } \left( \cfrac { { t }^{ -1/2 } }{ -1/2 }  \right) ={ \left( 1-{ x }^{ 2 } \right)  }^{ -1/2 }+c$$
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Single Correct Medium Published on 17th 09, 2020
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