Solve the integral thing 2

By letting $x$ to $-x$ , we get

$\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}{\dfrac{1}{(a+\sin{x})^{2}}dx}=2 \times \int_{0}^{\frac{\pi}{2}}{\frac{a^{2}+\sin^{2}{x}}{(a^{2}-\sin^{2}{x})^{2}}dx}$

Now, let $\displaystyle \tan{x}=t$ then,

$\displaystyle I=2 \times \int_{0}^{\infty}{\dfrac{(a^{2}+1)t^{2}+a^{2}}{((a^{2}-1)t^{2}+a^{2})^{2}}dt}$

$\displaystyle \int_{0}^{\infty}{\dfrac{(a^{2}+1)t^{2}+a^{2}}{((a^{2}-1)t^{2}+a^{2})^{2}}dt}=\dfrac{a^{2}+1}{a^{2}-1} \times \int_{0}^{\infty}{\dfrac{1}{((a^{2}-1)t^{2}+a^{2})}dt}+\dfrac{-2a^{2}}{a^{2}-1}\int_{0}^{\infty}{\dfrac{1}{((a^{2}-1)t^{2}+a^{2})^{2}}dt}$

Those are easy to integrate by trigonometric substitution, we get

$\displaystyle I=\dfrac{\pi a}{(a^{2}-1)^{3/2}}=\pi a$ so,

$\displaystyle a=\boxed{\sqrt{2}} \approx 1.41421$

We have $I_a \; = \; \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{(a+ \sin x)^2} \; = \; \tfrac12\int_{-\pi}^\pi \frac{dx}{(a + \sin x)^2}\; = \; 2i\int_{|z|=1} \frac{z\,dz}{(z^2 + 2iaz - 1)^2} \; = \; 2i\int_{|z|=1}\frac{z\,dz}{(z-\lambda_+)^2(z - \lambda_-)^2}$ where $\lambda_\pm \; = \; -ai \pm i\sqrt{a^2-1}$ Thus $I_a \; = \; -4\pi \mathrm{Res}_{z=\lambda_+} \frac{z}{(z-\lambda_+)^2(z - \lambda_-)^2} \; = \; -4\pi\left.\frac{d}{dz}\left(\frac{z}{(z-\lambda_-)^2}\right)\right|_{z=\lambda_+} \; = \; 4\pi\left. \frac{z+\lambda_-}{(z - \lambda_-)^3}\right|_{z = \lambda_+} \; = \; 4\pi\frac{\lambda_+ \lambda_-}{(\lambda_+ - \lambda_-)^3} \; = \; \frac{\pi a}{(a^2 - 1)^{\frac32}}$ for any $a >1$ , and hence $I = \pi a$ when $a^2 -1 = 1$ , and hence when $a = \boxed{\sqrt{2}}$ .