I am asked to show that $$\Gamma (x) \cos(ax) = b^x \int_{0}^{\infty} \mathrm{d} t \enspace t^{x-1} e^{-bt \cos(a)} \cos(bt \sin(a)).$$
A change of variables $t \to \frac{t}{b}$ shows that $b$ is superfluous and we can just eliminate it from the RHS. The exercise recommends that I write $\sin$ and $\cos$ using Euler's formula, and so the RHS becomes
$$ \frac{1}{2} \int_{0}^{\infty} \mathrm{d}t \enspace t^{x-1} e^{-t \cos(a)} \left[ e^{it \sin(a)} + e^{-it \sin(a)} \right] $$
and I want to get this to the form $$\frac{1}{2} \int_{0}^{\infty} \mathrm{d}t \enspace t^{x-1} e^{-t} \left[ e^{iax} + e^{-iax} \right] = \frac{1}{2} \int_{0}^{\infty} \mathrm{d}t \enspace t^{x-1} e^{-t \cos(a)} \left[ e^{iax + t \cos (a) - t} + e^{-iax + t \cos(a) -t} \right]$$
I have tried various things like performing $t \to \frac{t}{\cos(a)}$, $t \to t \tan(a)$, $t \to \frac{t}{\sin(a)}$. I have also tried integrating by parts the RHS (where I integrate $\cos \exp$) with no success. It seems there's some clever substitution that I'm just missing. I would appreciate any hints/guidance. Also, per Mathematica, the result I'm trying to show is indeed correct (given some constraints on $a$).
The trick in this case is to rewrite the cosine term as the real part of a single exponential function.
$$\cos(bt\sin a)=\Re(e^{ibt\sin a})$$
The integral becomes
$$\int_0^{+\infty}\mathrm dt\, t^{x-1}e^{-bt(\cos a-i\sin a)}=\int_0^{+\infty}\mathrm dt\, t^{x-1}\exp(-bte^{-ia})=b^{-x}e^{iax}\Gamma(x)$$
Considering only the real part, then $\Re(e^{iax})=\cos ax$, completing the proof.
$$\int_0^{+\infty}\mathrm dt\, t^{x-1} e^{-bt\cos a}\cos(bt\sin a)\color{blue}{=b^{-x}\Gamma(x)\cos ax}$$