I have this integral:

$$\int\sin^3t \, dt$$

I have tried partial integration with $\sin t \cdot \sin^2t$, but then I get another integral to evaluate which needs partial integration:

$$\dots \int \cos^2t \cdot \sin t \, dt \dots$$

Which gives me another integral that needs partial integration and so on. I get stuck in a partial integration loop.

How should I evaluate this?

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    $\begingroup$ You could do a substitution to solve the second integral, since the derivative of cosine is -sine. $\endgroup$ – Andrés E. Caicedo Jan 14 '12 at 0:21
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    $\begingroup$ Use "Show steps" button. wolframalpha.com/input/?i=%5Cint+%28sin+t%29%5E3+dt $\endgroup$ – Karatug Ozan Bircan Jan 14 '12 at 0:57
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    $\begingroup$ Using the result of this recent question, we have $$\sin^3 x=\frac{3}{4}\sin x - \frac{1}{4}\sin 3x$$ which is more easily integrated. $\endgroup$ – Dilip Sarwate Jan 14 '12 at 1:34
  • $\begingroup$ One method is to rewrite sin x as $(-i/2)(e^{ix}-e^{-ix})$. The integral then becomes one where it's obvious how to proceed, though messy. $\endgroup$ – Ben Crowell Jan 14 '12 at 1:44

$$\int \sin^{3}(t)dt=\int \sin(t)(1-\cos^{2}(t))dt$$ $$=\int \sin(t)dt-\int \sin(t)\cos^{2}(t)dt$$ $$=-\cos(t)-\int \sin(t)\cos^{2}(t)dt$$

Try $u=\cos(t)$ so that $du=-\sin(t)dt$ and the second integral is equivalent to

$-\int u^{2}du=-\frac{1}{3}u^{3}=-\frac{1}{3}\cos^{3}(t)$

Which gives us a final answer of

$$\int \sin^{3}(t)dt=-\cos(t)+\frac{1}{3}\cos^{3}(t)+C$$


The easiest way is to write your function as $(\sin t)(1-\cos^2 t)=\sin t -\sin t\; \cos^2 t$. For the second part, make the substitution $u=\cos t$.

A similar idea works for anything of the shape $(\sin^m t)(\cos^n t)$ where $m$ and $n$ are integers at least one of which is odd.

For example, to integrate $\sin^2 t \;\cos^5 t$, rewrite as $\sin^2 t \;\cos^4 t\; \cos t$. Then express $\cos^4 t$ as $(1-\sin^2 t)^2$, and make the substitution $u=\sin t$.

Comment: Your integration by parts would also have solved the problem. Let $u=\sin^2 t$ and $dv=\sin t\,dt$. Then $du=2\sin t\,\cos t$, and we can take $v=-\cos t$. So we have $$\int \sin^3 t\,dt =-\cos t \, \sin^2 t + \int 2\sin t\, \cos^2 t\, dt.$$ The remaining integral yields to the substitution $w=\cos t$.

But let's pretend we didn't notice this. In the integral, replace $\cos^2 t$ by $1-\sin^2 t$. After a little while, we arrive at $$\int \sin^3 t\,dt=-\cos t\, \sin^2 t +2\cos t -\int 2\sin^3 t\,dt.$$ Now it really looks as if we are stuck in the loop you referred to. The above equation seems to say "I will tell you the answer if you give me the answer." However, let $I=\int \sin^3 t\,dt$. Then we have $$I=-\cos t\,\sin^2 t +2\cos t -2I.$$ It follows that $3I=-\cos t\,\sin^2 t +2\cos t$. Now we can divide by $3$ to get $I$. Don't forget to add the arbitrary constant of integration.

The above trick comes up surprisingly often. When it looks as if we are going in circles, we usually are going in circles. But not always!

  • $\begingroup$ But then I have to integrate ${\sin t} \cdot {\cos^2t}$ which is the same thing. $\endgroup$ – Dimme Jan 14 '12 at 0:22
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    $\begingroup$ @Dimme: what do you think the derivative of $\cos^{3}(t)$ is? $\endgroup$ – Henry Jan 14 '12 at 0:32
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    $\begingroup$ No, it is easy. As I wrote, let $u=\cos t$. Then $du=-\sin t\, dt$, or $\sin t\, dt=du$. So you are integrating $-u^2\,du$. $\endgroup$ – André Nicolas Jan 14 '12 at 0:33

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