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How would I integrate the following:

.$$\int\frac{\sin^3x}{\cos x}\,\mathrm dx.$$

I am not sure what to do. I could split $\sin^3x=(1-\cos^2x)(\sin x)$

Then get $\int \tan(x)(1-\cos^2x)$

But would this be the right way to go.

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    $\begingroup$ Try the Maple command IntTutor which gives hints and shows the required solution step by step. $\endgroup$
    – user64494
    Commented Sep 20, 2013 at 16:51

3 Answers 3

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Let $u = \cos x \implies du = -\sin x\,dx$

$$\begin{align} \int \dfrac{\sin^3x}{\cos x} \,dx & = -\int \dfrac{-\sin x\sin^2 x}{\cos x}\,dx \\ \\ & = -\int \dfrac{-\sin x(1 - \cos^2 x)}{\cos x}\,dx \\ \\ & = -\int \dfrac{(1 - u^2)}{u}\,du \\ \\ & = \int u \,du - \int\dfrac 1u \,du\\ \\ & = \frac{u^2}{2} - \ln |u| + C \\ \\ & = \frac 12 \cos^2 x - \ln |\cos x| + C\end{align}$$

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  • $\begingroup$ I see so would I get $\int -\frac{1}{u}+u$ and then get ln(u)+u^2/2 $\endgroup$ Commented Sep 20, 2013 at 16:47
  • $\begingroup$ Almost: you get $-\ln u + \dfrac{u^2}2 + c$. Then replace $u = \cos x$, and you're done! $\endgroup$
    – amWhy
    Commented Sep 20, 2013 at 16:47
  • $\begingroup$ I see that make make sense. Ty $\endgroup$ Commented Sep 20, 2013 at 16:48
  • $\begingroup$ You're welcome! Just note the $-\ln u$. $\endgroup$
    – amWhy
    Commented Sep 20, 2013 at 16:49
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Actually you get $\int (\tan x-\sin x\cos x)dx$ which should be quite straight forward

$$\int\frac{\sin^3x}{\cos x}dx = \int\frac{\sin x(1-\cos^2x)}{\cos x}dx = \int (\tan x - \sin x\cos x) dx = \int (\tan x - \frac{1}{2}\sin 2x)dx$$

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  • $\begingroup$ Actually, $\tan x - \sin x \cos x = \tan x - \frac{1}{2}\sin(2x)$. $\endgroup$
    – J. J.
    Commented Sep 20, 2013 at 16:51
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    $\begingroup$ Not quite: $\int (\tan x - \sin x \cos x)\, dx$ $\endgroup$
    – amWhy
    Commented Sep 20, 2013 at 16:51
  • $\begingroup$ oh yeah, thanks $\endgroup$ Commented Sep 20, 2013 at 16:57
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Hint: Do the change $y=\cos x$.

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