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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.

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    $\begingroup$ $\sin^3 x= (1-\cos^2 x)\sin x $. Let $u=\cos x$ $\endgroup$ – David Mitra Aug 11 '15 at 14:26
  • $\begingroup$ @DavidMitra Done Thanks $\endgroup$ – Taylor Ted Aug 11 '15 at 14:28
  • $\begingroup$ @TaylorTed The trigonometric functions such as $\sin$ and $\cos$ must be uprighted. Please use the correct notations. $\endgroup$ – Zhanxiong Aug 11 '15 at 14:35
  • $\begingroup$ Do you Bioche's rules to determine the relevant change of variable? $\endgroup$ – Bernard Aug 11 '15 at 14:46
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$$\int \frac{\sin^3 x}{\cos^2 x}dx$$ $$=\int \frac{(1-\cos^2 x)\sin x}{\cos^2 x}dx$$ Let $\cos x=t\implies -\sin x dx=dt$ $$=-\int \frac{(1-t^2)dt}{t^2}$$ $$=-\int (t^{-2}-1)dt$$ $$=-\left(-\frac{1}{t}-t\right)+c$$ $$=\frac{1}{t}+t+c$$ $$=\cos x+\sec x+c$$

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

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$\bf{My\; Solution::}$ 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 $\displaystyle I = \frac{1}{\cos x}+\cos x+\mathcal{C}$

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