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. 
 A: Have a look here:
http://www.wolframalpha.com/input/?i=integrate+sin%282x%29+sin%28x%29+from+0+to+Pi
...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
A: 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.
A: 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 $$\int_{0}^{\pi} \sin{2x} \sin{x} \, dx = \left ( \frac{2}{3} \sin^3{x} \right )_{0}^{\pi} = 0 $$
A: 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.
A: Somewhat inspired by lhf's answer:
$$\int_0^\pi\sin\;2u\;\sin\;u\;\mathrm{d}u$$
$$\int_{0-\frac{\pi}{2}}^{\pi-\frac{\pi}{2}}\sin\left(2\left(u+\frac{\pi}{2}\right)\right)\;\sin\left(u+\frac{\pi}{2}\right)\mathrm{d}u$$
$$-\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u$$
$$-\left(\int_{-\frac{\pi}{2}}^0\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$
$$-\left(-\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u+\int_0^{\frac{\pi}{2}}\sin\;2u\;\cos\;u\;\mathrm{d}u\right)$$
and we see that the integral is zero.
It is convenient here that $\sin\;2u\;\cos\;u$ is an odd function.
A: 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 $f(\pi/2+x)=-f(\pi/2-x)$.
A: 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.
A: 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.
A: HINT: Use this formula  $$\cos C - \cos D = 2 \sin\frac{(C+D)}{2} \cdot \sin\frac{(D - C)}{2}$$
A: 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.
A: $∫\sin2x\sin x \,dx$
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)
$=(2/3)[0-0]=0$
