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.

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

$$\sin x=2\sin\frac x2\cos\frac x2=\frac{2u}{1+u^2}\\ \cos x=\cos^2\frac x2-\sin^2\frac x2=\frac{1-u^2}{1+u^2}$$


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

  • $\begingroup$ Yes, I got your result and then I tried another method.Thank you $\endgroup$
    – DaniVaja
    Feb 7 '20 at 15:49
  • $\begingroup$ There's an error in the denominator. $\endgroup$
    – Bernard
    Feb 7 '20 at 15:55
  • $\begingroup$ Sure about that? The rational function that I got was$$\frac{2-2u^2}{\left(u^2+1\right) \left(9 u^2+12 u+5\right)}.$$ $\endgroup$ Feb 7 '20 at 15:56
  • $\begingroup$ 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 $\endgroup$
    – DaniVaja
    Feb 7 '20 at 15:58
  • $\begingroup$ @Bernard Thanks, I do that quite often ... Fixed $\endgroup$
    – user170231
    Feb 7 '20 at 16:11

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