How to prove $ \int^{\pi}_{-\pi} \sin(ax)\sin(bx) = 0 $ Question:
I've been working on a problem involving inner product spaces with inner product given by:
$$ \langle f, g\rangle = \int^{\pi}_{-\pi} f(x)g(x)dx \hspace{0.5cm} \text{for every } f,g \in \mathcal{C}[-\pi, \pi] $$ As part of the problem, I've come to a point where the following inner product
$$ \int^{\pi}_{-\pi} \sin(ax)\sin(bx) = 0 \hspace{0.5cm} \text{ if } a \not = b \text{ and } a, b \in \mathbb{Z^{+}}$$ must be true for the purposes of the problem. 
Approach:
First I tried to simply evaluate the integral using integration by parts:
$$ -\frac{\sin(ax)\cos(bx)}{b}\bigg|^{\pi}_{-\pi} + \frac{a}{b}\int^{\pi}_{-\pi}\cos(ax)\cos(bx)dx $$
evaluating to
$$ 0 + \frac{a}{b}\int^{\pi}_{-\pi}\cos(ax)\cos(bx)dx $$
I took this another step arriving at:
$$ \frac{a}{b} \cdot \frac{\cos(ax)\sin(bx)}{b} + \frac{a^2}{b^2} \cdot \int^{\pi}_{-\pi} \sin(ax)\sin(bx)dx $$
$$ 0 + \frac{a^2}{b^2} \cdot \int^{\pi}_{-\pi} \sin(ax)\sin(bx)dx $$
At this point, it seems that the method I am using to evaluate the integral will lead to the integral repeating itself infinitely with a larger coefficient each time but still evaluating to zero each time. Is this enough to conclude that the integral evaluates to zero?
 A: We have 
\begin{align}
\int^{\pi}_{-\pi} \sin(ax)\sin(bx) \mathbb{d}x&=2\int^{\pi}_{0} \sin(ax)\sin(bx) \mathbb{d}x 
\end{align}
Because the function is even. 
\begin{align}
&=2 \left( \int^{\pi/2}_{0} \sin(ax)\sin(bx)\mathbb{d}x + \int^{\pi}_{\pi/2} \sin(ax)\sin(bx) \mathbb{d}x\right)\\
&=2 \left( \int^{\pi/2}_{0} \sin(a(\pi/2-x))\sin(b(\pi/2-x))\mathbb{d}x + \int^{\pi}_{\pi/2} \sin(ax)\sin(bx) \mathbb{d}x\right)\\
&\pi/2-x=u, \frac{du}{dx} = -1, \text{ bounds are now $\pi$ and $\pi/2$}\\
&=2 \left( -\int^{\pi}_{\pi/2} \sin(au)\sin(bu)\mathbb{d}u + \int^{\pi}_{\pi/2} \sin(ax)\sin(bx) \mathbb{d}x\right)\\
&=0
\end{align}
A: Alternatively, set $z=e^{i\theta}$ so that you get:
\begin{align}
\int^\pi_{-\pi} \sin(ax)\sin(bx) dx &=\int_{|z|=1} \frac{z^a-z^{-a}}{2i} \frac{z^b-z^{-b}}{2i} \frac{dz}{iz}\\
& = \frac{i}{4}\int_{|z|=1} z^{a+b-1} -z^{a-b-1} -z^{b-a-1}+z^{-a-b-1}dz
\end{align}
Since $b,a \in \mathbb{Z}^+$ and $a\neq b$ we have no $z$ with power $-1$. So we get:
\begin{align}
\int^\pi_{-\pi} \sin(ax)\sin(bx) dx = 0
\end{align} 
By the residue theorem.
A: $\cos x - \cos y =$ 
$-2\sin((x+y)/2)   \sin((x -y)/2);$  
$x,y \in \mathbb{R}.$
Set:
$az:= (x+y)/2$,  $bz:= (x -y)/2,$ :
$(1/2)[-\cos (a+b)z +\cos(a-b)z]$
$=\sin (az) \sin (bz)$.
Note: 
$a-b, a+b \in \mathbb{Z}$.
Since $ \cos$ is even,  we can choose 
$n=a+b, |a-b|$,  i.e.
$ n \in \mathbb{Z+}$, and get:
$\int_{0}^{π} cos(nz)dz = 0.$
$\rightarrow:$
$\int_{-π}^π [(-1/2)\cos (a+b)z +$
$(1/2)\cos(a-b)z]dz =0$ 
$\int_{-π}^π \sin(az) \sin(bz) dz $.
