How to find integrals of this form?

Question

How to calculate the integral of the form $$\int_0^x \sqrt{a+b \sin(ct)} \mathrm{dt}$$ where $a,b,c$ are constants? An integral of this form is to be solved while formulating something. Does the integral not converge for some values of the constants?

Answer 1

Assume $a \neq 0$:

$$\int_0^x \sqrt{a+b \sin(ct)} \mathrm{dt}=\sqrt{a} \int_0^x \sqrt{1+B \sin(ct)} \mathrm{dt}=$$

Where $B=b/a$. Similarly (as Jack D'Aurizio pointed out) we can get rid of $c$, assuming $c \neq 0$:

$$=\frac{\sqrt{a}}{c} \int_0^X \sqrt{1+B \sin(s)} \mathrm{ds}$$

$X=cx$. It is now enough to consider the integral with two parameters:

$$I(B,X)=\int_0^X \sqrt{1+B \sin(s)} \mathrm{ds}$$

To get this into the form consistent with elliptic integrals, we can replace:

$$\sin s=-1+2 \sin^2 \left( \frac{s}{2}+\frac{\pi}{4} \right)$$

Which makes the integral:

$$I(B,X)=\int_0^X \sqrt{1-B+2B \sin^2 \left( \frac{s}{2}+\frac{\pi}{4} \right)} \mathrm{ds}=$$

$$=2 \int_{\pi/4}^{X/2+\pi/4} \sqrt{1-B+2B \sin^2 (p)}~ \mathrm{dp}=2\sqrt{1-B} \int_{\pi/4}^{X/2+\pi/4} \sqrt{1-k^2 \sin^2 (p)}~ \mathrm{dp}$$

$$k^2=\frac{2B}{B-1}$$

Where we now have the well known form of the incomplete elliptic integral of the second kind (https://en.wikipedia.org/wiki/Elliptic_integral).

The parameter $k^2$ is chosen that way to get the canonical form. Note that either $\sqrt{1-B}$ or $k$ must be imaginary for real $B$.