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Solve $$ \int_0^\pi \frac {x} {1+\sin x} dx $$ using $\sin x = \frac {2\tan(x/2)} {1+\tan^2(x/2)}$ and the substitution $t = \tan(x/2)$. I tried doing this but I got to a point where my integral limits were $0$ to $\infty $. This happened when I substituted for $\tan(x/2)$ Is there a way of doing this using this substitution only? And also why does this happen?

UPDATE: What I did - $$ I = \int_0^\pi \frac {\pi-x} {1+\sin x} dx = \int_0^\pi \frac {\pi} {1+ \sin x} dx - I$$ Using $\sin x = \frac {2\tan(x/2)} {1+\tan^2(x/2)}$ $$ $$ $$ 2I = \int_0^\pi \frac {\pi} {1+\sin x} dx = \pi\int_0^\pi \frac {\sec^2(x/2)} {1+ \tan^2(x/2)+2 \tan(x/2)} dx$$ Now if there were no limits, this could've been solved easily by $t = \tan x$. But I can't do that because if I did, the limits would become $0$ to $\infty$. A way to solve this would be multiplying and dividing by $1+ \sin x$ but I don't want to do that. I want to use $t=\tan x$

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  • $\begingroup$ Maybe you should try integration by parts first to get rid of that free $x$ on the top there. $\endgroup$
    – Bob Jones
    Mar 17, 2017 at 8:00
  • $\begingroup$ Set $x\rightarrow x-\pi$ and see what happpens..nearly the same question was asked a few days ago by the user @bui so have a look in his profile $\endgroup$
    – tired
    Mar 17, 2017 at 8:08

4 Answers 4

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Hint. Alternatively, by the change of variable $u=\pi-x$, one gets $$ I=\int_0^\pi \frac {x} {1+\sin x} dx=\int_0^\pi \frac {\pi-u} {1+\sin (\pi-u)} du=\pi\int_0^\pi \frac {1} {1+\sin u} du-I $$ the latter integral being easier to evaluate.

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    $\begingroup$ Always the same old good trick ! $\endgroup$ Mar 17, 2017 at 8:10
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After the main symmetry trick, another symmetry trick and a rationalization: $$ \int_{0}^{\pi}\frac{du}{1+\sin u}=2\int_{0}^{\pi/2}\frac{1-\sin u}{\cos^2 u}\,du =2\left[\tan u-\frac{1}{\cos u}\right]_{0}^{\pi/2}=\color{red}{\large2}.$$

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  • $\begingroup$ I don't want to use rationalization, I want to do it using the substitution u = tanx. Tricks are fine but the substitution has to be that. $\endgroup$
    – Cowgirl
    Mar 17, 2017 at 9:38
  • $\begingroup$ @AnswerQuicklyPlease: you still have to split the integration range in halves since $\sin(x)$ is not injective on $(0,\pi)$. And after that, rationalization leads to the solution way faster than the given substitution. $\endgroup$ Mar 17, 2017 at 9:44
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If we must solve the problem with the suggested substitution, the integral becomes $\int_0^\infty\frac{4\arctan t dt}{(1+t)^2}$. The identity $\int_0^\infty f(t)dt=\int_0^1\left(f(t)+\tfrac{1}{t^2}f\left(\tfrac{1}{t}\right)\right)dt$ changes this to $\int_0^1\frac{2\pi dt}{(1+t)^2}=\pi$.

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Avoid the limits $(0, \infty)$ by reexpressing the anti-derivative without $ \tan \frac x2$, i.e

$$ \int\frac {1} {1+\sin x} dx = \int\frac {\sec^2\frac x2} {1+ \tan^2 \frac x2 +2 \tan \frac x2} dt\\ =-\frac {2} {1+ \tan \frac x2} = -\frac{2 \sin\frac x2}{\sin \frac x2+\cos \frac x2}$$

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