How to prove that $\int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}$ using real methods? Could you possibly help me prove $$\int_0^{\infty}\cos(\alpha t)e^{-\lambda t}\,\mathrm{d}t=\frac{\lambda}{\alpha^2+\lambda^2}$$
I'm an early calculus student, so I would appreciate a thorough answer using real and as basic as possible methods. Thank you in advance.

Note that
$$\frac d{dt}\frac{-\lambda\cos(\alpha t)e^{-\lambda t}+\alpha\sin(\alpha t)e^{-\lambda t}}{\alpha^2+\lambda^2}=\cos(\alpha t)e^{-\lambda t}$$
And apply the fundamental theorem of calculus.
