I need to integrate the following function:$$\frac{\cos(x)}{1+\sin(x)}\sqrt{\frac{2-\sin(x)}{2+\sin(x)}}$$
For this I use the substitution $2+\sin(x)=y^2$
I get $$2\int{\frac{\sqrt{4-y^2}}{y^2-1}} dy$$.
Then I use $y=2\sin(z)$.
So I get:
$$I=8\int\frac{{\cos^2(z)}}{4\sin^2(z)-1}dz$$
$$\implies I=2\left[\frac{\cos^2(z)}{\sin^2(z)-\frac{1}{4}}\right]$$
$$-2z+\frac{3}{2}\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz + C $$
After this, how to integrate $$\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz$$ (without using any hypergeometric functions) ? You may use logarithmic functions.
Well, multiply numerator and denominator by $4\csc^2(z)$:
$$\int\frac{1}{\sin^2(z)+\frac{1}{4}}\space\text{d}z=\int\frac{4\csc^2(z)}{4\csc^2(z)\left(\sin^2(z)+\frac{1}{4}\right)}\space\text{d}z=\int\frac{4\csc^2(z)}{\cot^2(z)+5}\space\text{d}z$$
Now, substitute $u=\cot(z)$:
$$\int\frac{4\csc^2(z)}{\cot^2(z)+5}\space\text{d}z=-4\int\frac{1}{u^2+5}\space\text{d}u=\text{C}-\frac{4\arctan\left(\frac{u}{\sqrt{5}}\right)}{\sqrt{5}}$$