So I just had an exam and one of the exercises was this integral:
$$ \int_{0}^{\pi}\frac{\sin(x)}{7+6\cos(x)-2\sin(x)}dx $$
My first thought was the the substitution $u=\tan(x/2)$.So I found $\sin x, \cos x$ and $dx$ and I replaced in my integral.Then I remembered that teacher said something that this substitution does not work in some cases. He said something about the range. So I give up on this method, then I split the integral from $0$ to $\pi/2$ and $\pi/2$ to $pi$. Now the first integral I tried to solve again with $u=\tan(x/2)$ and at the second one I got stuck.
Another method I tried is with King property: $f(a+b-x)=f(x)$ but I got nothing.
Other substitution I tried is $x=\pi-t$ but I also got stuck on this.
So my question is how to solve this integral?
I solved this kind of integral and the solving process was more "elegant".Like I apply $x=\pi-x$ then I write the integral in 2 different ways.Then I find the sum of them….and on and on.
Best Answer
$\tan\frac x2$ is continuous on $(0,\pi)$, so no need to split up the integral as you mention.
$$I=\int_0^\pi\frac{\sin x}{7+6\cos x-2\sin x}\,\mathrm dx$$
With $u=\tan\frac x2$, we get $\mathrm du=\frac12\sec^2\frac x2\,\mathrm dx$ and
$$\sin x=2\sin\frac x2\cos\frac x2=\frac{2u}{1+u^2}\\ \cos x=\cos^2\frac x2-\sin^2\frac x2=\frac{1-u^2}{1+u^2}$$
Then
$$I=\int_0^\infty\frac{\frac{2u}{1+u^2}}{7+\frac{6(1-u^2)}{1+u^2}-\frac{4u}{1+u^2}}\frac{2\,\mathrm du}{1+u^2}=\int_0^\infty\frac{4u}{(1+u^2)(13-4u+u^2)}\,\mathrm du$$