Evaluate $\int_{0}^{\pi} \frac{x\cdot dx}{1+\cos(\alpha)\cdot \sin(x)}$
$$I = \int_{0}^{\pi} \frac{x\cdot dx}{1+\cos(\alpha)\cdot \sin(x)} $$
$$\Rightarrow 2I = \pi\cdot \int_{0}^{\pi} \frac{dx}{1+\cos(\alpha)\cdot \sin(x)} = \frac{2\pi}{\cos(\alpha)}\cdot \int_{0}^{\pi/2} \frac{dx}{\sec(\alpha)+ \sin(x)}$$
$$\Rightarrow I =\frac{\pi}{\cos(\alpha)}\cdot \int_{0}^{\pi/2} \frac{dx}{\sec(\alpha)+ \sin(x)}$$
How can I proceed further with this integral? Basically $\sec(\alpha )$ is creating the problem here. I could integrate it if there was $1$ in place of $\sec(\alpha)$, but can't think of a way to proceed from here
$\sec(\alpha)$ is just a constant, and for any $A>1$ we have
$$ \int_{0}^{\pi/2}\frac{dx}{A+\sin(x)}=\int_{0}^{\pi/2}\frac{dx}{A+\cos(x)} = 2\int_{0}^{\pi/4}\frac{dx}{(A-1)+2\cos^2(x)}$$ where the substitution $x=\arctan t$ turns the last integral into $$ 2\int_{0}^{1}\frac{dt}{(A-1)t^2+(A+1)}=\frac{1}{\sqrt{A^2-1}}\,\arctan\sqrt{\frac{A-1}{A+1}}. $$