Show that $\int_{-\infty}^{\infty} \frac{\cos (\theta) e^{\theta y}}{2 \cosh(\pi y/2)} y dy=\tan (\theta)$. I'm not a hardcore mathematician. I have exhausted all integration trick I learned in calculus including change of variables, integration by parts. That cosh on denominator is just very much in the way. How one should approach an integral like this. 

Here is a real method.
Let
$$I = \int_{-\infty}^\infty \frac{\cos \theta \, e^{\theta y} \, y}{2 \cosh \left (\frac{\pi y}{2} \right )} \, dy,$$
where $\theta \in (-\pi/2, \pi/2)$.
Now
$$I = \frac{\cos \theta}{2} \int_{-\infty}^0 \frac{y e^{\theta y}}{\cosh \left (\frac{\pi y}{2} \right )} \, dy + \frac{\cos \theta}{2} \int_0^\infty \frac{y e^{\theta y}}{\cosh \left (\frac{\pi y}{2} \right )} \, dy = I_1 + I_2.$$
In the first of the integrals, enforcing a substitution of $y \mapsto -y$ gives
\begin{align}
I_1 &= -\frac{\cos \theta}{2} \int_0^\infty \frac{y e^{-\theta y}}{\cosh \left (\frac{\pi y}{2} \right )} \, dy\\
&= -\cos \theta \int_0^\infty \frac{y e^{-\theta y}}{e^{\frac{\pi y}{2}} + e^{-\frac{\pi y}{2}}} \, dy \tag1\\
&= -\cos \theta \int_0^\infty \frac{y e^{-\theta y - \pi y/2}}{1 + e^{-\pi y}} \, dy\\
&= -\cos \theta \sum_{n = 0}^\infty (-1)^n \int_0^\infty y e^{-(n \pi + \pi/2 + \theta)} \, dy \tag2\\
&= -\cos \theta \sum_{n = 0}^\infty \frac{(-1)^n}{(n \pi + \pi/2 + \theta)^2} \tag3\\
&= -\frac{\cos \theta}{\pi^2} \sum_{n = 0}^\infty \frac{(-1)^n}{(n + \frac{1}{2} + \frac{\theta}{\pi})^2}\\
&= -\frac{\cos \theta}{\pi^2} \sum_{\substack{n = 0\\n \in \text{even}}}^\infty \frac{1}{\left (n + \frac{1}{2} + \frac{\theta}{\pi} \right )^2} - \frac{\cos \theta}{\pi^2} \sum_{\substack{n = 0\\n \in \text{odd}}}^\infty \frac{1}{\left (n + \frac{1}{2} + \frac{\theta}{\pi} \right )^2}\\ 
&= -\frac{\cos \theta}{4 \pi^2} \sum_{n = 0}^\infty \frac{1}{\left (n + \frac{1}{4} + \frac{\theta}{2 \pi} \right )^2} - \frac{\cos \theta}{4 \pi^2} \sum_{n = 0}^\infty \frac{1}{\left (n + \frac{3}{4} + \frac{\theta}{2 \pi} \right )^2} \tag4\\
&= -\frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left (\frac{\theta}{2 \pi} + \frac{1}{4} \right ) - \psi^{(1)} \left (\frac{\theta}{2 \pi} + \frac{3}{4} \right ) \right ] \tag5 
\end{align}
In a similar manner, for the second of the integrals it can be shown that
$$I_2 = \frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left (\frac{1}{4} - \frac{\theta}{2 \pi} \right ) - \psi^{(1)} \left (\frac{3}{4} - \frac{\theta}{2 \pi} \right ) \right ].$$
Thus
\begin{align}
I &= -\frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left [1 - \left (\frac{3}{4} - \frac{\theta}{2 \pi} \right ) \right ] + \psi^{(1)} \left (\frac{3}{4} - \frac{\theta}{2 \pi} \right ) \right ]\\
& \qquad \quad + \frac{\cos \theta}{4 \pi^2} \left [\psi^{(1)} \left [1 - \left (\frac{1}{4} - \frac{\theta}{2 \pi} \right ) \right ] + \psi^{(1)} \left (\frac{1}{4} - \frac{\theta}{2 \pi} \right ) \right ]\\
&= -\frac{\cos \theta}{4 \pi^2} \left [\pi^2 \operatorname{cosec}^2 \left (\frac{3 \pi}{4} - \frac{\theta}{2} \right ) - \pi^2 \operatorname{cosec}^2 \left (\frac{\pi}{4} - \frac{\theta}{2} \right ) \right ] \tag6\\
&= -\frac{\cos \theta}{4} \left (\frac{2}{\sin \theta + 1} + \frac{2}{\sin \theta - 1} \right )\\
&= -\frac{\cos \theta}{4} \cdot -\frac{4 \sin \theta}{\cos^2 \theta},
\end{align}
or finally
$$\int_{-\infty}^\infty \frac{\cos \theta \, e^{\theta y} \, y}{2 \cosh \left (\frac{\pi y}{2} \right )} \, dy = \tan \theta,$$
as required.

Explanation
(1) Using $\cosh x = (e^x + e^{-x})/2$.
(2) Geometric expansion.
(3) Integration by parts twice.
(4) Shifting of the indices. Here $n \mapsto 2n$ for $n$ even and $n \mapsto 2n + 1$ for $n$ odd.
(5) Using the series representation for the polygamma function.
(6) Using the reflexion relation for the trigamma function.
