Mathematics

Find $$\displaystyle \int \dfrac {1}{\sin x\cos^{3}x}dx.$$


SOLUTION
$$\int { \dfrac { 1 }{ \sin { x } \cos ^{ 3 }{ x }  }  } dx$$
$$=\int { \dfrac { 1 }{ \dfrac { \sin { x }  }{ \cos { x }  } .\cos { x } .\cos ^{ 3 }{ x }  }  } dx$$
$$=\int { \dfrac { 1 }{ \tan { x } \cos ^{ 4 }{ x }  }  } dx$$
$$=\int { \dfrac { \sec ^{ 4 }{ x }  }{ \tan { x }  }  } dx$$       $$\left( \because \cos x=\dfrac{1}{\sec x}\right)$$
Now, substitute $$t=\tan x$$
$$\Rightarrow \dfrac{dt}{dx}=\sec^2 x$$
$$\Rightarrow dt=\sec^2 xdx$$
          $$=\int { \dfrac { \sec ^{ 2 }{ x } \sec ^{ 2 }{ x }  }{ \tan { x }  }  } dx$$
          $$=\int { \dfrac { \left( 1+\tan ^{ 2 }{ x }  \right) \sec ^{ 2 }{ x }  }{ \tan { x }  }  } dx$$      $$ \left( \because \sec^2 x-\tan^x=1 \rightarrow \sec^2 x=1+\tan^2 x\right)$$
          $$=\int { \dfrac { \left( 1+t^2 \right)  }{ t }  } dx$$
          $$=\int \dfrac{1}{t}dt +\int t dt$$
          $$=ln\;t+\dfrac{t^2}{2}+c$$
          $$=ln \left| \tan^{-1}x\right| +\dfrac{\left( \tan^{-1}x\right)^2}{2}+c$$
Hence, the answer is $$ln \left| \tan^{-1}x\right| +\dfrac{\left( \tan^{-1}x\right)^2}{2}+c.$$

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Subjective Hard Published on 17th 09, 2020
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