TITLE: Evaluating $\int_0^{\pi/2}\sin(2nx)\sinh(a\sin x)\sin(a\cos x)dx $
QUESTION [6 upvotes]: I want to prove that 
$$\int_0^{\pi/2}\sin(2nx)\sinh(a\sin x)\sin(a\cos x)dx =\frac{(-1)^{n+1}\pi a^{2n}}{4(2n)!}$$
any one have an idea

REPLY [11 votes]: Note first that
$$\cos{\left ( a e^{-i x}\right)} = \cos{(a \cos{x})} \cosh{(a \sin{x})} + i \sin{(a \cos{x})} \sinh{(a \sin{x})}$$
so that the integral is 
$$\Im{\left [\int_0^{\pi/2} dx \, \sin{2 n x} \; \cos{\left ( a e^{-i x}\right)} \right ]}$$
Expand the cosine in a Taylor series, so that the integral is
$$\Im{\left [\sum_{k=0}^{\infty} \frac{(-1)^k a^{2 k}}{(2 k)!} \int_0^{\pi/2} dx \, \sin{2 n x} \;e^{-i 2 k x} \right ]}$$
The integral is zero unless $k=n$, and only the imaginary part is nonzero.  The result follows from the fact that
$$\int_0^{\pi/2} dx \, \sin^2{2 n x} = \frac{\pi}{4}$$
when $n \in \mathbb{N}$.