# Integral $\int_{0}^{\pi}\frac{\sin(x)}{7+6\cos(x)-2\sin(x)}dx$

$$\int_{0}^{\pi}\frac{\sin(x)}{7+6\cos(x)-2\sin(x)}dx$$

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.

• Here, Bioche's rules say you indeed should use the substitution $t=\tan\frac x2$, $dx=\frac{2\,dt}{1+t^2}$. Feb 7 '20 at 16:00
• what is king property .it isn't written in my textbook by 'king' name Feb 7 '20 at 17:08

$$\tan\frac x2$$ 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

$$I=\int_0^\infty\frac{\frac{2u}{1+u^2}}{7+\frac{6(1-u^2)}{1+u^2}-\frac{4u}{1+u^2}}\frac{2\,\mathrm du}{1+u^2}=\int_0^\infty\frac{4u}{(1+u^2)(13-4u+u^2)}\,\mathrm du$$

• Yes, I got your result and then I tried another method.Thank you Feb 7 '20 at 15:49
• There's an error in the denominator. Feb 7 '20 at 15:55
• Sure about that? The rational function that I got was$$\frac{2-2u^2}{\left(u^2+1\right) \left(9 u^2+12 u+5\right)}.$$ Feb 7 '20 at 15:56
• I don't even remember but I'm pretty sure that I got the correct rational function.I used half angle formula, I got a rational function then I tried another method.Anyway, it's easy from here to solve it.Thanks again Feb 7 '20 at 15:58
• @Bernard Thanks, I do that quite often ... Fixed Feb 7 '20 at 16:11