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.
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.
HINT: Use this formula $$\cos C - \cos D = 2 \sin\frac{(C+D)}{2} \cdot \sin\frac{(D - C)}{2}$$
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.
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
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.
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 $$\int_{0}^{\pi} \sin{2x} \sin{x} \, dx = \left ( \frac{2}{3} \sin^3{x} \right )_{0}^{\pi} = 0 $$
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.
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.
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.
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 $f(\pi/2+x)=-f(\pi/2-x)$.
$∫\sin2x\sin x \,dx$
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)
$=(2/3)[0-0]=0$
