Is there any short method to solve the integral $I=\int {{\sin3x}\cdot\sin{5x\over2}\over\sin {x\over2}}dx$ Given integral:$$I=\int\,\,{{\sin3x}\cdot\sin{5x\over2}\over\sin {x\over2}}\,dx$$
I have solved the given integral but it's too lengthy and tiresome to write here.Also is there any method to solve the integral with help of Complex Integration
(I'm new here on math.SE and can't type frequently so for me it'll take too much time to type the complete solution here since I'm using iPad)
In case anyone wants to see my solution, I'll attach the image of paper-solution here
Sorry for not being supportive, need your help!
 A: $$\frac{\sin3x\sin\frac{5}{2}x}{\sin{\frac{x}{2}}}=\frac{2\sin\frac{x}{2}\cos\frac{x}{2}(3-4\sin^2x)\sin\frac{5}{2}x}{\sin{\frac{x}{2}}}=$$
$$=(\sin2x+\sin3x)(1+2\cos2x)=\sin2x+\sin3x+\sin4x+\sin5x+\sin{x}$$
A: Using $\sin\theta=\frac{e^{i\theta}-e^{i\theta}}{2i}$, one has
\begin{eqnarray}
&&{{\sin3x}\cdot\sin{5x\over2}\over\sin {x\over2}}\\
&=&\frac{1}{2i}{{(e^{3ix}-e^{-3ix})}\cdot(e^{5ix/2}-e^{-5ix/2})\over e^{ix/2}-e^{-ix/2}}\\
&=&\frac{1}{2i}{{(e^{3ix}-e^{-3ix})}\cdot(e^{3ix}-e^{-2ix})\over e^{ix}-1}\\
&=&\frac{1}{2ie^{5ix}}{{(e^{6ix}-1)}(e^{5ix}-1)\over e^{ix}-1}\\
&=&\frac{1}{2ie^{5ix}}(e^{6ix}-1)(e^{4ix}+e^{3ix}+e^{2ix}+e^{ix}+1)\\
&=&\frac{1}{2ie^{5ix}}(e^{10ix}+e^{9ix}+e^{8ix}+e^{7ix}+e^{6ix}-e^{4ix}-e^{3ix}-e^{2ix}-e^{ix}-1)\\
&=&\frac{1}{2i}(e^{5ix}+e^{4ix}+e^{3ix}+e^{2ix}+e^{ix}-e^{-ix}-e^{-2ix}-e^{-3ix}-e^{-4ix}-e^{-5ix})\\
&=&\sin(5x)+\sin(4x)+\sin(3x)+\sin(2x)+\sin x.
\end{eqnarray}
A: Set $y = x/2$ and then use the identity
$$\sin(5y) = \sin y(2\cos 2y + 2 \cos 4y + 1)$$
Then you will essentially have to compute
$$ \int \sin(6y)(2\cos 2y + 2 \cos 4y + 1) \text{d}y$$
Now you can use the identity $$2 \sin a \cos b = \sin(a+b) + \sin(a-b)$$
The rest if left as an exercide.
For completeness: to prove the identity:
$$\sin(5y) = \sin y(2\cos 2y + 2 \cos 4y + 1)$$
Consider
$$\sin 5x - \sin x = 2 \cos 3x \sin 2x$$
(using $\sin (a+b) - \sin (a-b) = 2 \cos a \sin b$)
$$ 2 \cos 3x \sin 2x = 4 \cos 3x \cos x \sin x$$
(using $\sin 2x = 2 \sin x \cos x$)
$$ = 2\sin x (\cos 4x + \cos 2x)$$
Using $\cos (a+b) + \cos (a-b) = 2 \cos a \cos b$
A: Hint: 
If $f(m, n)=\dfrac{\sin mx\sin nx}{\sin\dfrac x2}, $
$f(m, n)-f(m, n-1)=2\sin mx\cos\dfrac{(2n-1)x}2=? $( Use Werner's formula)
$$\implies\int f(m,n)dx=\int f(m,n-1)dx+\int\left(\sin\dfrac{2m+2n-1}2+\sin\dfrac{2m-2n+1}2\right)dx$$
Set $m=\dfrac52,\dfrac32$ and add
Now $\displaystyle\int f\left(m,\dfrac12\right)dx=?$
Finally set $m=3$
