Mathematics

Evaluate $$\int {\dfrac{{{{\cos }^2}x}}{{{{\sin }^3}x{{\left( {{{\sin }^5}x + {{\cos }^5}x} \right)}^{\frac{3}{5}}}}}dx} $$


SOLUTION
$$ I=\int { \dfrac { { { { \cos   }^{ 4 } }x } }{ { { { \sin   }^{ 3 } }x{ { \left( { { { \sin   }^{ 5 } }x+{ { \cos   }^{ 5 } }x } \right)  }^{ \frac { 3 }{ 5 }  } } } } dx }  $$
$$I =\int { \dfrac { { { { \sec   }^{ 2 } }xdx } }{ { { { \tan   }^{ 6 } }x{ { \left( { 1+\frac { 1 }{ { { { \tan   }^{ 5 } }x } }  } \right)  }^{ \frac { 3 }{ 5 }  } } } }  } $$

Let 
$$ \tan  x=p\, \, \, \, { \sec ^{ 2 }  }xdx=dp $$

Therefore,
$$ \int { \dfrac { { dp } }{ { { p^{ 6 } }{ { \left( { 1+\frac { 1 }{ { { p^{ 5 } } } }  } \right)  }^{ \frac { 3 }{ 5 }  } } } }  }  \\ 1+\dfrac { 1 }{ { { p^{ 5 } } } } =k \\ \dfrac { { dk } }{ { dp } } =\dfrac { { -5 } }{ { { p^{ 6 } } } }  \\$$

Therefore,
$$ I=-\dfrac { { 1 } }{ 5 } \int { { { \left( k \right)  }^{ \frac { { -3 } }{ 5 }  } }dk }  \\ =\dfrac { { -1 } }{ 5 } \left( { \dfrac { 5 }{ 2 }  } \right) { k^{ \frac { 2 }{ 5 }  } }+C \\ =\dfrac { { -1 } }{ 2 } { \left[ { \dfrac { { { p^{ 5 } }+1 } }{ { { p^{ 5 } } } }  } \right] ^{ \frac { 2 }{ 5 }  } } \\ =-\dfrac { 1 }{ 2 } { \left[ { \dfrac { { { { \tan   }^{ 5 } }x+1 } }{ { { { \tan   }^{ 5 } }x } }  } \right] ^{ \frac { 2 }{ 5 }  } }+C \\ =-\dfrac { 1 }{ 2 } { \left( { 1+{ { \cot   }^{ 5 } }x } \right) ^{ \frac { 2 }{ 5 }  } }+C$$

Hence, this is the answer.
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Subjective Medium Published on 17th 09, 2020
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