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| tan(x)*x dx = | x*tan(x) dx
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0 0
$${{i\,\pi^2}\over{24}}-{{2\,i\,\arctan \left({{\sin 2}\over{\cos 2+1
}}\right)+\log \left(2\,\cos 2+2\right)-i\,{\it li}_{2}(-e^{2\,i})-i
}\over{2}}$$
Ответ (Неопределённый)
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| tan(x)*x dx = C + | x*tan(x) dx
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$$-{{x\,\log \left(\sin ^2\left(2\,x\right)+\cos ^2\left(2\,x\right)+
2\,\cos \left(2\,x\right)+1\right)+2\,i\,x\,{\rm atan2}\left(\sin
\left(2\,x\right) , \cos \left(2\,x\right)+1\right)-i\,{\it li}_{2}(
-e^{2\,i\,x})-i\,x^2}\over{2}}$$