Evaluating $\int _0^{\pi }\frac{x}{1+\cos \theta \cdot \sin x} \ \mathrm{d}x$ for $\theta \in (0, \pi)$

Question

The question is as follows:

Let $\theta \in (0, \pi)$ and $I=\int _0^{\pi }\frac{x}{1+\cos \theta \cdot \sin x} \ \mathrm{d}x$

Remark: $\forall \theta \in (0, \pi) \wedge \forall x \in [0, \pi]$ we have $1+\cos \theta \cdot \sin x >0$

a) With the help of a $x=\pi-t$ change of variable, show that: $$I=\frac{\pi}{2}\int _0^{\pi }\frac{1}{1+\cos \theta \cdot \sin x} \ \mathrm{d}x$$

b) Show that: $$I=\pi \int _0^{\frac{\pi}{2} }\frac{1}{1+\cos \theta \cdot \cos x} \ \mathrm{d}x$$

c) Show that: $$I=\frac{\pi \theta}{\sin{\theta}}$$

I would talk about what I have done, but I wasn't able to do much other than:

\begin{align} \int _0^{\pi }\frac{x}{1+\cos \theta \cdot \sin x} \ \mathrm{d}x &= -\int _0^{\pi }\frac{\pi-t}{1+\cos \theta \cdot \sin{(\pi-t)}} \ \mathrm{d}t \\\\ &= -\int _0^{\pi }\frac{\pi-t}{1+\cos \theta \cdot \sin{t}} \ \mathrm{d}t \\\\ &= \int_0^{\pi}\frac{t}{1+\cos \theta \cdot \sin{t}} \ \mathrm{d}t-\int _0^{\pi }\frac{\pi}{1+\cos \theta \cdot \sin{t}} \ \mathrm{d}t \\ \end{align}

Any help would be appreciated, thanks.

Answer 1

Hint This integral $I=\int_{0}^{\pi}\frac{\pi-t}{1+\sin t\cos\theta}\text{d}t$ needs to be rewritten in terms of x so we say $t=x$ (notice the lack of a negative, you did not rewrite your bounds correctly). Now we need to add our 2 different forms of $I$. So $2I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x\cos\theta}\text{d}x+\int _0^{\pi }\frac{x}{1+\cos \theta \cdot \sin x} \ \mathrm{d}x$ and then from there we can get parts b and c