# $\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$

$$\int_ \frac\pi6^\frac\pi2 \frac {3dx}{2\sin2x+1}$$

I tried using this substitute: $$(\tan x=u),(sin2x= \frac{2\tan x}{1+\tan^2x} =\frac{2u}{u^2+1}), (dx=\frac{1} {u^2+1}du)$$ and after long answer I get:

$$\frac{6+3\sqrt{3}}{6+4\sqrt{3}}.\ln(\frac{\tan x+2 -\sqrt{3}}{\tan x+2+\sqrt{3}})$$ and now I can not use $$F(\frac\pi2)-F(\frac\pi6)$$ because $$\tan(\frac\pi2)=\infty$$

Since $$\lim_{x\to\frac{\pi}{2}}\frac{\tan x +2-\sqrt{3}}{\tan x +2+\sqrt{3}}=\lim_{x\to\frac{\pi}{2}}\frac{1+\frac{2-\sqrt{3}}{\tan x}}{1+\frac{2+\sqrt{3}}{\tan x}}=1.$$ Then you can get what you want.
I would do $$u=\tan x$$ too, and then your integral becomes$$\int_{1/\sqrt3}^\infty\frac3{t^2+4t+1}\,\mathrm dt.$$But$$\int\frac3{t^2+4t+1}\,\mathrm dt=\frac{\sqrt3}2\left(\log \left(-t+\sqrt{3}-2\right)-\log\left(t+\sqrt{3}+2\right)\right)$$and therefore$$\begin{multline}\int_{1/\sqrt3}^\infty\frac3{t^2+4t+1}\,\mathrm dt=\\=\lim_{t\to\infty}\frac{\sqrt3}2\left(\log\left(-t+\sqrt{3}-2\right)-\log\left(t+\sqrt{3}+2\right)\right)-\\-\frac{\sqrt3}2\left(\log\left(-\frac1{\sqrt3}+\sqrt{3}-2\right)-\log\left(\frac1{\sqrt3}+\sqrt{3}+2\right)\right)=\\=\frac{\sqrt3}2\log\left(\frac{5+3\sqrt3}2\right).\end{multline}$$
• How?: $$\int\frac3{t^2+4t+1}\,\mathrm dt=\frac{\log \left(-t+\sqrt{3}-2\right)-\log\left(t+\sqrt{3}+2\right)}{2\sqrt3}$$ I used Partial Fractions in This situation Oct 24, 2019 at 14:44
• It follows from$$\frac3{t^2+4t+1}=\frac3{\left(t+2-\sqrt3\right)\left(t-2-\sqrt3\right)}=\frac{3/4}{t-2-\sqrt3}-\frac{3/4}{t+2-\sqrt3}.$$ Oct 24, 2019 at 14:48
• It should be $$\frac3{\left(t+2+\sqrt3\right)\left(t+2-\sqrt3\right)}$$ Oct 24, 2019 at 14:51
• Actually, it should be$$\frac3{t^2+4t+1}=\frac{\sqrt{3}}{2\left(t-\sqrt{3}+2\right)}-\frac{\sqrt{3}}{2\left(t+\sqrt{3}+2\right)}.$$I've edited my answer. Oct 24, 2019 at 14:57