How do I integrate $\frac {\sin^3x}{\cos^2x}$ How do I integrate $\frac {\sin^{3}x}{\cos^{2}x}$. I have tried to convert to $\tan$, but I could not reach to conclusion. Any help will be appreciated.
Thanks.

$$\int { \frac { \sin ^{ 3 }{ x }  }{ \cos ^{ 2 }{ x }  } dx=\int { \frac { \sin { x } \cdot \sin ^{ 2 }{ x }  }{ \cos ^{ 2 }{ x }  } dx=-\int { \frac { 1-\cos ^{ 2 }{ x }  }{ \cos ^{ 2 }{ x }  } d\cos { x }  }  }  } =\\ =\int { d\left( \cos { x }  \right)  } -\int { \frac { 1 }{ \cos ^{ 2 }{ x }  } d\left( \cos { x }  \right) =\cos { x } +\frac { 1 }{ \cos { x }  } +C }  $$
