Mathematics

The value of $$\int \dfrac {1}{\sqrt {\sin^{3}x \cos^{5}x}} dx$$ is


ANSWER

$$\dfrac {-2}{\sqrt {\tan x}} + \dfrac {2}{3} (\tan x)^{3/2} + C$$


SOLUTION
Let $$I=\int { \cfrac { 1 }{ \sqrt { \sin ^{ 3 }{ x } \cos ^{ 5 }{ x }  }  } dx } $$
$$I=\int { \cfrac { 1 }{ \sqrt { \cfrac { \sin ^{ 3 }{ x }  }{ \cos ^{ 3 }{ x }  } \cos ^{ 8 }{ x }  }  } dx } =\cfrac { 1 }{ \cos ^{ 4 }{ x } \sqrt { \tan ^{ 3 }{ x }  }  } dx$$
$$I=\int { \cfrac { \sec ^{ 4 }{ x }  }{ \sqrt { \tan ^{ 3 }{ x }  }  } dx } =\int { \cfrac { \sec ^{ 3 }{ x } .\sec ^{ 2 }{ x }  }{ \sqrt { \tan ^{ 3 }{ x }  }  } dx } $$
$$I=\int { \cfrac { \left( 1+\tan ^{ 2 }{ x }  \right)  }{ \sqrt { \tan ^{ 3 }{ x }  }  } \sec ^{ 2 }{ x } dx } $$
Let $$\tan { x } =t$$
$$\tan { x } =t$$
$$\tan { x } =t$$
$$\sec ^{ 2 }{ x } dx=dt$$
So, $$I=\int { \cfrac { 1+{ t }^{ 2 } }{ \sqrt { { t }^{ 3 } }  } dt } =\int { \cfrac { 1+{ t }^{ 2 } }{ { t }^{ 3/2 } } dt } $$
$$I=\int { { t }^{ -3/2 }dt } +\int { { t }^{ 1/2 }dt } $$
$$I=\cfrac { { t }^{ -1/2 } }{ -1/2 } +\cfrac { { t }^{ 3/2 } }{ 3/2 } +c$$
So, $$I=\cfrac { -2 }{ \sqrt { \tan { x }  }  } +\cfrac { 2 }{ 3 } { \left( \tan { x }  \right)  }^{ 3/2 }+c$$
where C is arbitary constant
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Single Correct Medium Published on 17th 09, 2020
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