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$$\int_0^\pi \frac{x}{1+\sin \alpha \sin x} dx$$ I need some Hints about how to begin with the problem, because I can't think of anything

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    $\begingroup$ You can get rid of the $x$ in the numerator by doing $x\mapsto \pi-x$ (this symmetry is asked for many times on this site). Then, you will essentially have $\int 1/(1+a\sin x)\,dx$, another integral asked for many times on this site. $\endgroup$
    – mickep
    Nov 21, 2015 at 10:49

2 Answers 2

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HINT:

  1. Use the fact that $$\int^{\pi}_{0}f(x)dx=\int^{\pi}_{0}f(\pi+0-x)dx$$ where $f(x)=\frac{x}{1+\sin \alpha \sin x}$

AND

  1. Use the substitution $u=\tan \frac{x}{2}$ after converting $\sin x$ to $\tan \frac{x}{2}$ in the expression $$\int^{\pi}_{0}\frac{1}{1+\sin \alpha \sin x} dx$$
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Put $t=\pi-x$ , then you will get

$$I=\int_0^\pi \frac{x}{1+\sin\alpha\cdot\sin x} dx=\int_0^\pi \frac{\pi -x}{1+\sin\alpha \cdot \sin(\pi-x)} dx$$

$$\Rightarrow 2I=\pi\int_0^\pi \frac{1}{1+\sin\alpha\cdot\sin x} dx$$

Now put $T=\tan\left(\frac{x}{2}\right)$.

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    $\begingroup$ Note that you can / should use a slash before common functions to get $\sin$ instead of $sin$, and that multiplication can be written using \cdot rather than a period. $\endgroup$ Nov 21, 2015 at 11:03

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