Is this integral correct $\int_{-\pi/2}^{\pi/2}\cos^2(x)\sin[\alpha+\beta\tan( x)]\mathrm dx?$ How can we show that: 

$$\int_{-\pi/2}^{\pi/2}\cos^2(x)\sin[\alpha+\beta\tan( x)]\mathrm dx=\frac{1+\beta}{2}\cdot\frac{\pi}{e^{\beta}}\cdot\sin(\alpha)$$

assume $\alpha$ and $\beta$ are real numbers.
I am not sure how to begin to tackle this problem. Any help, please!
 A: We can simplify the integral, by taking into account the parity:
\begin{align}
 I&=\int_{-\pi/2}^{\pi/2}\cos^2(x)\sin[\alpha+\beta\tan( x)]\mathrm dx\\
 &=\sin \alpha\int_{-\pi/2}^{\pi/2}\cos^2(x)\cos\left( \beta\tan x \right)\,dx
\end{align}
Then, using the substitution $t=\tan x$,
\begin{equation}
 I=\sin \alpha\int_{-\infty}^\infty\frac{\cos\beta t}{\left( 1+t^2 \right)^2}\,dt
\end{equation} 
Introducing a parameter $a>0$,
\begin{align}
 J(a)&=\sin \alpha\int_{-\infty}^\infty\frac{\cos\beta t}{\left( a+t^2 \right)^2}\,dt\\
 I&=J(1)
\end{align} 
One may express
\begin{align}
 J(a)=-\sin \alpha\frac{d}{da}\int_{-\infty}^\infty\frac{\cos\beta t}{ a+t^2 }\,dt\\
\end{align}
The integral to be evaluated is a classical Fourier transform (see for example here):
\begin{equation}
 J(a)=-\sin \alpha\frac{d}{da}\frac{\pi}{\sqrt{a}}e^{-\left|\beta\right|\sqrt{a}}
\end{equation} 
The derivative gives directly
\begin{equation}
 J(a)=\sin \alpha \frac{\pi}{2a^{3/2}}\left( 1+\left|\beta\right|\sqrt{a} \right)e^{-\left|\beta\right|\sqrt{a}}
\end{equation} 
and thus
\begin{equation}
 I=\frac\pi 2 \sin \alpha\left( 1+\left|\beta\right| \right)e^{-\left|\beta\right|}
\end{equation} 
A: Exploiting symmetry reveals
$$\int_{-\pi/2}^{\pi/2} \cos^2(x) \sin(\alpha+\beta \tan(x))\,dx=\sin(\alpha)\int_{-\pi/2}^{\pi/2} \cos^2(x) \cos(\beta \tan(x))\,dx\tag1$$
Next, enforcing the substitution $x\mapsto \arctan(x)$ in the right-hand side of $(1)$, we find that for $\beta>0$
$$\begin{align}
\int_{-\pi/2}^{\pi/2} \cos^2(x) \cos(\beta \tan(x))\,dx&=\int_{-\infty}^\infty \frac{\cos(\beta x)}{(1+x^2)^2}\,dx\\\\
&=\int_{-\infty}^\infty \frac{e^{i\beta x}}{(1+x^2)^2}\,dx\\\\
&=2\pi i \text{Res}\left(\frac{e^{i\beta z}}{(1+z^2)^2}, z=i\right)\\\\
&= 2\pi i \left.\left(\frac{d}{dz}\frac{e^{i\beta z}}{(z+i)^2}\right)\right|_{z=i}\\\\
&=2\pi i \left(-\frac i4 (1+\beta)e^{-\beta}\right)\\\\
&=\frac{\pi}{2}(1+\beta)e^{-\beta}\tag2
\end{align}$$
Putting together $(1)$ ands $(2)$, and exploiting the evenness  in $\beta$ of the result, yields the coveted result
$$\int_{-\pi/2}^{\pi/2} \cos^2(x) \sin(\alpha+\beta \tan(x))\,dx=\frac{\pi}{2}\sin(\alpha)(1+|\beta|)e^{-|\beta|}$$
