# How to solve $\int_0^{\pi} \sin{2x}\sin{x} dx$?

How to solve $$\int_0^{\pi} \sin{2x}\sin{x} dx$$

Edit: Sorry! I should have described more. This is not a homework. Recently, Out of the blue I got interest in physics and started reading and solving problems. This is part of a physics problem where I got stuck (because I forgot all high school formulae.). Thanks all of you guys for wonderful solutions.

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Heck!! Why is it not formatted? –  claws Oct 21 '10 at 8:49
Also, this seems like it might be a homework question. If you are asking for help with your homework, please add the tag "homework" to your question, and edit the question to mention what you have already tried. –  Rahul Oct 21 '10 at 8:55
No, its not a homework question. Its just that I forgot most of my maths. Trying to revise. –  claws Oct 21 '10 at 10:28
@claws: If it is not a homework question, please clarify it. This question is formulated like one. People may not provide a complete answer if they suspect it is a homework. @Chandru1: Please don't add the homework tag unless it really is homework. –  KennyTM Oct 21 '10 at 17:29
Claws - it's blue, not blew! Like the sky (en.wikipedia.org/wiki/Out_of_the_blue_(idiom)) –  George Lowther Oct 21 '10 at 19:05

There is no need for trigonometric identities, complex exponentials or the like. Observe that \begin{eqnarray} \int_{0}^{\pi} \sin(2 x) \sin (x) dx = \int_{0}^{\pi/2} \sin(2 x) \sin (x) dx + \int_{\pi/2}^{\pi} \sin(2 x) \sin (x) dx \end{eqnarray} By a change of variables ($x \to \pi - x$) and the oddness of the integrand on the interval $[0,\pi]$, you find that \begin{eqnarray} \int_{\pi/2}^{\pi} \sin(2 x) \sin (x) dx = - \int_{0}^{\pi/2} \sin(2 x) \sin (x) dx , \end{eqnarray} which implies that the original integral vanishes identically.

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Plotting $\sin{2x}\sin{x}$ in $[0,\pi]$ shows that it's symmetrical with respect to $x=\pi/2$ and so the integral is zero.

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HINT: Use this formula $$\cos C - \cos D = 2 \sin\frac{(C+D)}{2} \cdot \sin\frac{(D - C)}{2}$$

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This formula has quite an impressive name, by the way: mathworld.wolfram.com/ProsthaphaeresisFormulas.html –  Hans Lundmark Oct 21 '10 at 13:09
Certainly not! I wasn't the one who downvoted. –  Hans Lundmark Oct 21 '10 at 15:10
@hans: I am not blaming you hans! Downvoting without a reason is upsetting! –  anonymous Oct 21 '10 at 15:12
Well, whoever downvoted you did it before I wrote my comment, so I don't think it had anything to do with not mentioning the name of the formula. –  Hans Lundmark Oct 21 '10 at 16:10
Another thing: I didn't even know that this formula had a name until long after I got my PhD in mathematics, so don't feel too bad if you didn't learn the name in high scool. :-) –  Hans Lundmark Oct 21 '10 at 16:12

I'd suggest using an identity for $\sin 2x$ that rewrites it in terms of $\sin x$ and $\cos x$, then using substitution ($u=\sin x$) on the result.

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Two more ways: use Euler's formula $\sin ax=(e^{iax}-e^{-iax})/2i$, or integrate by parts twice to get an equality where your sought integral appears twice and can be solved for.

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Or simply note that $\sin 2x$ and $\sin x$ are eigenfunctions with different eigenvalues to the Sturm–Liouville problem $-f''=\lambda f$, $f(0)=f(\pi)=0$, and therefore automatically orthogonal with respect to the $L^2$ inner product on the interval $[0,\pi]$. ;-) –  Hans Lundmark Oct 21 '10 at 9:23
Lundmark - hehe. That's great. You seen this? mathoverflow.net/questions/42512/… –  George Lowther Oct 21 '10 at 19:01
@George: Yes, that's a funny thread, although I don't think this example is quite as farfetched as many of the things mentioned there. Integrals like this appear all the time when you (for example) solve the heat equation on an interval by separation of variables and Fourier series, and it's very convenient to know that they are automatically zero. –  Hans Lundmark Oct 21 '10 at 19:38

...and for all the steps of the derivation (click on "show steps"):
http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29

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If you use the identity $\sin{2x} = 2 \cos{x}\sin{x}$ then you get

$$\int \sin{2x} \sin{x} \, dx = 2 \int \sin^2{x} \cos{x} \, dx$$

and now you can make the substitution $u = \sin{x}$ to get

$$2 \int u^2 \, du = \frac{2}{3} u^3 + C = \frac{2}{3} \sin^3{x} + C$$

Therefore $$\int_{0}^{\pi} \sin{2x} \sin{x} \, dx = \left ( \frac{2}{3} \sin^3{x} \right )_{0}^{\pi} = 0$$

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Writing $\sin x^n$ instead of $\sin^n x$ or even $(\sin x)^n$ is confusing. –  Aryabhata Oct 21 '10 at 14:31
Yes, I'm sorry, I just scrolled down and I totally missed Isaac's answer. I tried to delete my post, but it only said that it was a vote to delete it. –  Adrián Barquero Oct 21 '10 at 14:33

$$\int_0^\pi\sin\;2u\;\sin\;u\;\mathrm{d}u$$

$$\int_{0-\frac{\pi}{2}}^{\pi-\frac{\pi}{2}}\sin\left(2\left(u+\frac{\pi}{2}\right)\right)\;\sin\left(u+\frac{\pi}{2}\right)\mathrm{d}u$$

$$-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u$$

$$-\left(\int_{-\frac{\pi}{2}}^0\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$

$$-\left(-\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$

and we see that the integral is zero.

It is convenient here that $\sin\;2u\;\cos\;u$ is an odd function.

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yeah. why is it zero? I mean we were calculating area right? –  claws Oct 21 '10 at 11:44
@claws, you should see this. –  Ｊ. Ｍ. Oct 21 '10 at 12:14
Or perhaps more simply: $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ –  Aryabhata Oct 21 '10 at 14:19
I should note that this method is quite general, in that it only exploits the symmetry (i.e. if the function is odd or even) of the integrand. For instance, this trick works for showing that $\int_{\mathbb{R}}t^{2j+1}\exp\left(-t^{2k}\right)\mathrm{d}t$ for $j$ and $k$ positive integers is 0. –  Ｊ. Ｍ. Oct 22 '10 at 0:59

To supplement lhf's observation of the symmetry of the graph, note that if $f(x)=\sin(2x)\sin(x)$, then the symmetry observed is that $f(\pi/2+x)=-f(\pi/2-x)$, which means that the integral over $[0,\pi/2]$ exactly cancels the integral over $[\pi/2,\pi]$.

The symmetry can be seen algebraically by noting that $\sin(\pi/2+x)=\sin(\pi/2-x)$, and that $$\sin(2(\pi/2+x))=\sin(\pi+2x)=-\sin(2x)=\sin(-2x)=-\sin(\pi-2x)=-\sin(2(\pi/2-x)),$$ with each equality being evident from the unit circle definition of $\sin$. Multiplying yields $f(\pi/2+x)=-f(\pi/2-x)$.

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So we are trying to integrate the following expression $~~~\rightarrow ~~~ \displaystyle\int_0^{\pi} \sin 2x\sin x\ dx$.

To do the this, we will need to make an appropriate substitution inside of the integrand and would be nice to use the double-angle trigonometric identity for $\sin 2x$. Doing this leads us to the following:

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\displaystyle\int_0^{\pi} \sin 2x\sin x\ dx$

$$\Rightarrow~\displaystyle\int_0^{\pi} 2\sin x \cos x \sin x\ dx~~~~~\Big(\because~\sin 2x =2\sin x \cos x \Big)$$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle\int_0^{\pi} 2\sin^{2} x \cos x\ dx$

Let: $~u =\sin x$

$du= \cos x\ dx$

$dx=\dfrac{1}{\cos x}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle 2\int_0^{\pi} u^{2} \cos x \cdot \dfrac{1}{\cos x}\ du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle 2\int_0^{\pi} u^{2}~du$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle 2\Bigg[~ \dfrac{u^{3}}{3} \Bigg|_{x=0}^{x=\pi} ~~\Bigg]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle 2\Bigg[~ \dfrac{\sin^{3} x}{3} \Bigg|_{x=0}^{x=\pi} ~~\Bigg]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle 2\Bigg[~ \dfrac{(\sin (\pi))^{3}}{3}-\dfrac{(\sin (0))^{3}}{3} ~~\Bigg]$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~\Rightarrow~\displaystyle 2\Big[~0 ~\Big]~~~~\Big(\because~ \sin (\pi)=0 ~~\&~~ \sin (0)=0\Big)$

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=~~0$

$$\therefore~\displaystyle\int_0^{\pi} \sin 2x\sin x\ dx=0$$

HINT: Another way to look at it is that $\sin (x)$ is zero on the unit circle at coordinates 0 and $\pi$, so once you get to the line where it says $u^{2}$ and you have seen that the $\cos (x)$ has cancelled and realized that your substitution you made was $\sin (x)$ and that is what you must plug back into $u$ to put the answer back in terms of x or another option would be to change the limits of integration with the u - substitution. But once you see that u =$\sin (x)$ and notice the limits of integration are exactly where $\sin (x)=0$, you can quickly say that the integral is $0$ without even integrating using the power rule and evaluating the end-points.

Okay, I hope that this has helped out. Let me know if there is anything point covered that did not make much sense for doing.

Thanks.

Good Luck.

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$∫\sin2x\sin x \,dx$

NOTE: $\sin2x=2\sin x\cos x$

$\int 2\sin x \cos x \sin x \,dx$

$∫2\sin x\sin x \cos x \,dx$

$=2∫\sin x\sin x\cos x\,dx$

$=2∫(\sin x)^2d(\sin x)$ [Differentiation of $\sin x$: $d(\sin x)=\cos x\,dx$

$=(2/3)\left[(\sin x)^3\right]_0^\pi$ ($\pi=$180 Degree)

$=(2/3)[0-0]=0$

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I have edited you post using MathJax - it is definitely more readable now. For some basic information about writing math at this site see e.g. here, here, here and here. You can also learn by example by viewing source code by others (click on edit or right click and select Show Math As/TeX Commands.) –  Martin Sleziak Apr 14 at 11:30