Using integration by parts show that: $$\int_0^{+\infty} e^{-x} \cos x\, dx = \int_0^{+\infty} e^{-x} \sin x\, dx$$

So far I have just expanded out, and I don't see an end, is there a trick I am missing?
A rather simple and detailed approach :

$$\int_0^\infty e^{-x}\cos(x)\mathrm{d}x = \int_0^\infty e^{-x}(\sin(x))'\mathrm{d}x = \big[ e^{-x}\sin(x)\big]_0^\infty - \int_0^\infty(e^{-x})'\sin(x)\mathrm{d}x$$

$$=$$

$$\lim_{x\to \infty} \big[ e^{-x}\sin(x)\big] - 0 - \int_0^\infty-e^{-x}\sin(x)\mathrm{d}x $$

$$=$$

$$\boxed{\int_0^\infty e^{-x}\sin(x)\mathrm{d}x}$$
apply integration by part twice then you'll get the same integral with extra terms then call integral I then it is a first order linear equation so, solve it for I.
Other approach add 
$$\int e^{-x} i \cos(x)$$ to original integral. So desired result is real part of  $$\int_{0}^{\infty} e^{-x}(\sin(x)+ i \cos(x))dx = \int e^{(-1+i)x} dx$$.