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This is related to my previous asked question here. The integration can be simplified as (As pointed out by Sivaram, the integral diverges to $-\infty$ as $c\to1+$):

$$\int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin y+c}}\ \text{d}y,\ \ c>1.$$

I am appreciated if there is a specific answer (it would be great if there are some details).


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For $c=1$, the integral diverges. You might want to change it to $c>1$ – user17762 Mar 13 '11 at 3:14
Really? I believe that $c=1$ is the easiest case to compute and converges to 2. – Pengfei Mar 13 '11 at 3:24
1… – user17762 Mar 13 '11 at 3:27
Sorry I must be wrong. It should always negative. – Pengfei Mar 13 '11 at 3:49
up vote 0 down vote accepted

Using formula 288.03 in Byrd and Friedman's handbook:

$$\int_{\varphi}^{\pi/2}\frac{\sin^m u}{\sqrt{a+b\sin\;u}}\mathrm du=\frac2{\sqrt{a+b}}\int_0^{F\left(\arcsin\left(\sqrt{\frac{1-\sin\;\varphi}{2}}\right)\mid\frac{2b}{a+b}\right)}\left(1-2\mathrm{sn}^2\left(u\mid\frac{2b}{a+b}\right)\right)^m\mathrm du$$

for $a > b > 0$, where $F(\phi|m)$ is the incomplete elliptic integral of the first kind and $\mathrm{sn}(u|m)$ is a Jacobian elliptic function, we have

$$\int_{-\pi/2}^{\pi/2} \frac{\sin y}{\sqrt{\sin y+c}}\mathrm{d}y=2\left(\sqrt{1+c}\;E\left(\frac2{1+c}\right)-\frac{c}{\sqrt{1+c}}K\left(\frac2{1+c}\right)\right)$$

where $K(m)$ is the complete elliptic integral of the first kind, and $E(m)$ is the complete elliptic integral of the second kind.

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