Mathematics

$$\displaystyle\int {{{dx} \over {\sqrt {{{\sin }^3}x\cos x} }} = g(x)}  + c \Rightarrow g(x) = $$


ANSWER

$$\displaystyle{{ - 2} \over {\sqrt {\tan x} }}$$


SOLUTION
Solution:
$$I=\displaystyle \int { \dfrac{ 1 }{ \sqrt { \sin ^{ 3 }{ x } \cos { x }  }  } dx } $$

$$=\displaystyle \int { \dfrac{ 1 }{ \dfrac{ \sqrt { \sin ^{ 3 }{ x }  }  }{ \sqrt { \cos ^{ 3 }{ x }  }  } \times \sqrt { \cos ^{ 3 }{ x }  } \times \sqrt { \cos { x }  }  } dx } $$

$$=\displaystyle \int { \dfrac{ \sec ^{ 2 }{ x } .dx }{ \sqrt { \tan ^{ 3 }{ x }  }  }  } $$      Let $$\tan x=t$$

$$\Rightarrow \sec^2x dx=dt$$

$$=\displaystyle \int { \dfrac{ dt }{ { t }^{ 3/2 } }  } $$

$$=-2t^{-1/2}+C$$

$$=\dfrac {-2}{\sqrt {\tan x}}+C=g(x)+C$$

$$\Rightarrow \ g(x)=\dfrac {-2}{\sqrt {\tan x}}$$
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Single Correct Medium Published on 17th 09, 2020
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