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

Question

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!

Answer 1

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

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

Answer 2

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}