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.
 A: $$\int \frac{\sin^3 x}{\cos^2 x}dx$$
$$=\int \frac{(1-\cos^2 x)\sin x}{\cos^2 x}dx$$
Let $\cos x=t\implies -\sin x dx=dt$
$$=-\int \frac{(1-t^2)dt}{t^2}$$
$$=-\int (t^{-2}-1)dt$$
$$=-\left(-\frac{1}{t}-t\right)+c$$
$$=\frac{1}{t}+t+c$$
$$=\cos x+\sec x+c$$
A: $$\int { \frac { \sin ^{ 3 }{ x }  }{ \cos ^{ 2 }{ x }  } dx=\int { \frac { \sin { x } \cdot \sin ^{ 2 }{ x }  }{ \cos ^{ 2 }{ x }  } dx=-\int { \frac { 1-\cos ^{ 2 }{ x }  }{ \cos ^{ 2 }{ x }  } d\cos { x }  }  }  } =\\ =\int { d\left( \cos { x }  \right)  } -\int { \frac { 1 }{ \cos ^{ 2 }{ x }  } d\left( \cos { x }  \right) =\cos { x } +\frac { 1 }{ \cos { x }  } +C }  $$
A: $\bf{My\; Solution::}$ Let $\displaystyle \int\frac{\sin^3 x}{\cos^2 x}dx  = \int\frac{\sin^2 x \cdot \sin x}{\cos^2 x}dx = \int\frac{(1-\cos^2 x)\cdot \sin x}{\cos^2 x}dx $
Now Let $\cos x = t\;,$ Then $\sin xdx = -dt$
So Integral $\displaystyle I = -\int\frac{1-t^2}{t^2}dt = -\int \frac{1}{t^2}dt+\int 1\cdot dt=\frac{1}{t}+t+\mathcal{C}$
So $\displaystyle I = \frac{1}{\cos x}+\cos x+\mathcal{C}$
