Ответ (Неопределённый)
[src]
$$-{{2\,x\,\log \left(\sin ^2\left(2\,x\right)+\cos ^2\left(2\,x
\right)+2\,\cos \left(2\,x\right)+1\right)-2\,x\,\log \left(\sin ^2
\left(2\,x\right)+\cos ^2\left(2\,x\right)-2\,\cos \left(2\,x\right)
+1\right)+4\,i\,x\,{\rm atan2}\left(\sin \left(2\,x\right) , \cos
\left(2\,x\right)+1\right)+4\,i\,x\,{\rm atan2}\left(\sin \left(2\,x
\right) , 1-\cos \left(2\,x\right)\right)+2\,i\,{\it li}_{2}(e^{2\,i
\,x})-2\,i\,{\it li}_{2}(-e^{2\,i\,x})}\over{8}}$$
1
/
|
| x
| -------- dx
| sin(2*x)
|
/
0
$$-{{i\,\arctan \left({{\sin 2}\over{\cos 2+1}}\right)}\over{2}}+{{i
\,\arctan \left({{\sin 2}\over{\cos 2-1}}\right)}\over{2}}-{{\log
\left(2\,\cos 2+2\right)}\over{4}}+{{\log \left(2-2\,\cos 2\right)
}\over{4}}-{{i\,{\it li}_{2}(e^{2\,i})}\over{4}}+{{i\,{\it li}_{2}(-
e^{2\,i})}\over{4}}+{{i\,\pi^2}\over{16}}$$
=
1
/
|
| x
| -------- dx
| sin(2*x)
|
/
0
$$\int\limits_{0}^{1} \frac{x}{\sin{\left(2 x \right)}}\, dx$$