$\int_{0}^{\pi} \frac{R^2 + R}{R^2 - 5} e^{-R \sin{\theta}} \ d \theta$ How Can I calculate this integral? 
$$\int_{0}^{\pi} \frac{R^2 + R}{R^2 - 5} e^{-R \sin {\theta}} \ d \theta , \ \ \ \theta \in ]0,  \pi[ \ \ \ , R \to \infty $$
I have tried evaluating this using dominating convergence theorem. My question is there any other way to evaluate this ?
Thank you. 
 A: Here are two slightly different approaches:


*

*As pointed out by @Zaid Alyafeai, you can write


$$ 0
\leq \int_{0}^{\pi} e^{-R\sin\theta} \,d\theta
= 2\int_{0}^{\pi/2} e^{-R\sin\theta} \,d\theta
\leq 2\int_{0}^{\pi/2} e^{-\frac{2R}{\pi}\theta} \,d\theta
\leq \frac{\pi}{R} \xrightarrow[R\to\infty]{}0. $$


*Good ol' truncation argument works here: for any $\epsilon \in (0,\frac{\pi}{2})$, we have
\begin{align*}
\int_{0}^{\pi/2} e^{-R\sin\theta} \,d\theta
&\leq \int_{0}^{\epsilon} e^{-R\sin\theta} \,d\theta + \int_{\epsilon}^{\pi/2} e^{-R\sin\theta} \,d\theta \\
&\leq \int_{0}^{\epsilon} \,d\theta + \int_{\epsilon}^{\pi/2} e^{-R\sin\epsilon} \,d\theta \\
&\leq \epsilon + \pi e^{-R\sin\epsilon}.
\end{align*}
You can pick $\epsilon = R^{-1/2}$ to show the convergence.
A: You seek the solution to
$$\int_{0}^{\pi} \frac{R^2 + R}{R^2 - 5} e^{-R \sin {\theta}} \ d \theta , \ \ \ \theta \in [0,  \pi], \ \ \ R \to \infty$$
We can dispatch the $R^2$ terms by noting that the ratio goes to unity as $R \to \infty$. For the remaining integral, let $x=\sin\theta$, so that
$$
dx=\cos\theta\ d\theta\\
d\theta=\frac{dx}{\sqrt{1-x^2}}
$$
and the integral can be written as
$$2\int_0^1\frac{e^{-Rx}}{\sqrt{1-x^2}}dx=\pi\left(\text{I}_0(R)-\text{L}_0(R) \right)$$
where $\text{I}_0$ and $\text{L}_0$ are the modified Bessel function of the first kind and the modified Struve function, respectively. (I'm not that smart, this result is from WolframAlpha.) What you really have to know is what the integrand looks like when you plot it up. It's quite apparent that
$$2\int_0^1\frac{e^{-Rx}}{\sqrt{1-x^2}}dx=0 \text{ as } R\to\infty$$
And therefore the solution you seek is
$$
\lim_{R\to\infty} \int_{0}^{\pi} \frac{R^2 + R}{R^2 - 5} e^{-R \sin {\theta}} \ d \theta=0
$$
A: Your integral is
$$\frac {R^2+R}{ (R^2-5)}\frac {1-e^{-R\pi}}{R}$$
the limit is $0$.
A: Hint: the indefinite integral is given by $$-\frac{(R+1)e^{-R\theta}}{R^2-5}+C$$
