<s>[INST] Evaluate $ \int \frac{1}{\sin{x}\cos^{3}x}dx $ $$ \int \frac{1}{\sin{x}\cos^{3}x}dx $$
$$\Rightarrow \int \frac{1}{\sin{x}\cos^{3}x}{\cos{x}\over \cos{x}}dx$$
$$\Rightarrow \int \frac{\sec^{4}{x}}{\tan{x}}dx$$
$$\Rightarrow \int \frac{\sec^{2}(1+\tan^{2}x)}{\tan{x}}dx$$
$$Substitution \tan{x}=t \Rightarrow \sec^{2}xdx=dt$$
$$\Rightarrow \int \frac{1+t^{2}}{t}dt$$
$$\Rightarrow \int \frac{1}{t}dt+\int tdt $$
$$\Rightarrow \log|t|+\frac{t^{2}}{2}+C$$
$$\Rightarrow \log|\tan{x}|+\frac{1}{2}\tan^{2}x+C$$
Any other elegant way to do this?
 [/INST] Another way, I am not sure it is an elegant way but is different than yours
$$\int  \frac { 1 }{ \sin { x } \cos ^{ 3 } x } dx=\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x }  }  }{ \sin { x } \cos ^{ 3 } x } dx= } \int { \frac { \sin { x }  }{ \cos ^{ 3 }{ x }  }  } dx+\int { \frac { 1 }{ \sin { x\cos { x }  }  } dx } =\\ =-\int { \frac { d\cos { x }  }{ \cos ^{ 3 }{ x }  } +\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x }  }  }{ \sin { x } \cos { x }  }  }  } dx=\frac { 1 }{ 2\cos ^{ 2 }{ x }  } +\int { \frac { \sin { x }  }{ \cos { x }  } dx+\int { \frac { \cos { x }  }{ \sin { x }  } dx } = } \\ =\frac { 1 }{ 2\cos ^{ 2 }{ x }  } -\ln { \left| \cos { x }  \right| +\ln { \left| \sin { x }  \right|  } +C= } \frac { 1 }{ 2\cos ^{ 2 }{ x }  } +\ln { \left| \tan { x }  \right|  } +C$$
</s>