Mathematics

$$\displaystyle \int \dfrac{\sin^{2} x \cos^{2} x}{(\sin^{2}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x)^{2}} dx$$ is equal to:


SOLUTION
$$\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 3 }{ x }  }{ { \left( \sin ^{ 2 }{ x } +\cos ^{ 3 }{ x } .\sin ^{ 2 }{ x } +\sin ^{ 3 }{ x } \cos ^{ 2 }{ x } +\cos ^{ 5 }{ x }  \right)  }^{ 2 } } dx } $$
$$=\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 2 }{ x }  }{ { \left\{ \left[ \sin ^{ 2 }{ x } \left[ \sin ^{ 3 }{ x } .\cos ^{ 3 }{ x }  \right]  \right] +\cos ^{ 2 }{ x } \left[ \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x }  \right]  \right\}  }^{ 2 } } dx } $$
$$\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 2 }{ x }  }{ { \left\{ \left( \sin ^{ 2 }{ x } +\cos ^{ 2 }{ x }  \right) \left[ \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x }  \right]  \right\}  }^{ 2 } } dx } $$
$$=\int { \cfrac { \sin ^{ 2 }{ x } .\cos ^{ 2 }{ x }  }{ { \left( \sin ^{ 3 }{ x } +\cos ^{ 3 }{ x }  \right)  }^{ 2 } } dx } $$
Divide $$\cos ^{ 3 }{ x } $$ in numerator and denominator
$$=\int { \cfrac { \sec ^{ 3 }{ x } .\tan ^{ 2 }{ x }  }{ { \left( \tan ^{ 3 }{ x } +1 \right)  }^{ 2 } } dx } $$
Let $$1+\tan ^{ 3 }{ x } =t$$
$$3\tan ^{ 2 }{ x } \sec ^{ 2 }{ x } dx=dt$$
$$\therefore \int { \cfrac { 1 }{ 3 } \cfrac { dt }{ { t }^{ 2 } }  } =\cfrac { -1 }{ 3 } \cfrac { 1 }{ t } +c$$
$$=\cfrac { 1 }{ 3\left( \tan ^{ 3 }{ x } +1 \right)  } +c$$
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Subjective Medium Published on 17th 09, 2020
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