Mathematics

Solve:$$\int \displaystyle \dfrac{\sin 2x}{\sin^4x+ \cos^4 x}dx$$


SOLUTION
Given the integral,
$$\int { \dfrac { \sin { 2x }  }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x }  }  } dx\\ =\int { \dfrac { 2\sin { x } \cos { x }  }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x }  }  } dx\\ =\int { \dfrac { \dfrac { 2\sin { x } \cos { x }  }{ \cos ^{ 4 }{ x }  }  }{ \dfrac { \sin ^{ 4 }{ x }  }{ \cos ^{ 4 }{ x }  } +\dfrac { \cos ^{ 4 }{ x }  }{ \cos ^{ 4 }{ x }  }  }  } dx\\ =\int { \sec ^{ 2 }{ x } \dfrac { 2\tan { x }  }{ \tan ^{ 4 }{ x } +1 }  } dx$$
Let us assume,
$$u=\tan { x } \\ \Rightarrow \dfrac { du }{ dx } =\tan { x } \\ \Rightarrow du=\tan { x } dx$$
Substituting the values of $$u$$ and $$du$$ we get,
$$\int { \sec ^{ 2 }{ x } \dfrac { 2\tan { x }  }{ \tan ^{ 4 }{ x } +1 }  } dx\\ =\int { \dfrac { 2u }{ { u }^{ 4 }+1 }  } du$$
Again let us assume,
$$v={ u }^{ 2 }\\ \Rightarrow \dfrac { dv }{ du } =2u\\ \Rightarrow dv=2udu$$
Substituting the values of $$v$$ and $$dv$$ we get,
$$\int { \dfrac { 2u }{ { u }^{ 4 }+1 }  } du\\ =\int { \dfrac { 1 }{ { v }^{ 2 }+1 }  } dv\\ =\tan ^{ -1 }{ v } +C\\ =\tan ^{ -1 }{ { u }^{ 2 } } +C\\ =\tan ^{ -1 }{ (\tan ^{ 2 }{ x } ) } +C\\ \therefore \int { \dfrac { \sin { 2x }  }{ \sin ^{ 4 }{ x } +\cos ^{ 4 }{ x }  }  } dx=\tan ^{ -1 }{ (\tan ^{ 2 }{ x } ) } +C.$$
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Subjective Medium Published on 17th 09, 2020
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