I am stuck with the following integral while studying some cases of anharmonic Oscillations:
$$\int_0^t\frac{\exp(-\mu t')}{\cos^2(\omega t'-\theta_0)}\, dt'$$
where, $\mu$, $\theta_0$ and $\omega$ have been considered as Real constants.
I have put this integral in Mathematica, which shows a very complicated result involving Hypergeometric functions, but I could not understand how the derivation can be done.
Would you please suggest me any relevant sources or method such that I can do this integral.
Let $ u = wt'-\theta_0 \Rightarrow dt' = \frac{du}{w} $, $ t' = \frac{(u+\theta_0)}{w}$ $a = w\times0-\theta_0$ and $b=wt-\theta_0$ Also set $\psi = \frac\mu{w}$ $$ w^{-1}\int_a^b e^{-\psi(u+\theta_0)}\sec^2(u)du=w^{-1}e^{-\psi\theta_0}\int_a^b e^{-\psi u}\sec^2(u)du=w^{-1}e^{-\psi\theta_0}\left(\left.e^{-\psi u}\tan{(u)}\right.\rvert_a^b+\psi\int_a^b e^{-\psi u}\tan{(u)} du\right)$$
Now to evaluate $$ \int_a^b e^{-\psi u}\tan{(u)} du $$ Set $ \xi = e^{-\psi u} \Rightarrow du = -\psi e^{\psi u} d\xi $ and $u = \frac{-\ln(r)}{\psi}$ So $c=e^{-\psi a}$ $ d =e^{-\psi b}$ $$ \int_a^b e^{-\psi u}\tan{(u)} du=\psi\int_c^d \tan{\left(\frac{\ln(\xi)}{\psi}\right)} d\xi=\psi i^{-1}\int_c^d \frac{e^{\frac{\ln(\xi)i}{\psi}}-e^{-\frac{\ln(\xi)i}{\psi}}}{e^{\frac{-\ln(\xi)i}{\psi}}+e^{\frac{\ln(\xi)i}{\psi}}} d\xi=\psi i^{-1}\left(\xi\rvert_c^d-2\int_c^d \frac{e^{-\frac{\ln(\xi)i}{\psi}}}{e^{\frac{-\ln(\xi)i}{\psi}}+e^{\frac{\ln(\xi)i}{\psi}}} d\xi\right)=\psi i^{-1}\left(\xi\rvert_c^d-2\int_c^d \frac{1}{(e^{\frac{-\ln(\xi)i}{\psi}}+e^{\frac{\ln(\xi)i}{\psi}})e^{\frac{\ln(\xi)i}{\psi}}} d\xi\right)=\psi i^{-1}\left(\xi\rvert_c^d-2\int_c^d \frac{1}{(1+e^{\frac{2\ln(\xi)i}{\psi}})} d\xi\right)=\psi i^{-1}\left(\xi\rvert_c^d-2\int_c^d \frac{1}{(1+\xi^{\frac{2i}{\psi}})} d\xi\right) $$ Overall $$ w^{-1}e^{-\psi\theta_0}\left(\left.e^{-\psi u}\tan{(u)}\right.\rvert_a^b+\psi^2i^{-1}\left(\xi\rvert_c^d-2\int_c^d \frac{1}{(1+\xi^{\frac{2i}{\psi}})} d\xi\right)\right) $$