Интеграл x*sec(x)*tan(x) (dx)

Вы ввели

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$$\int_{0}^{1} x \sec{\left (x \right )} \tan{\left (x \right )}\, dx$$

Ответ

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 |  x*sec(x)*tan(x) dx =  |  x*sec(x)*tan(x) dx
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$$-{{\sin ^22\,\log \left(\sin ^21+2\,\sin 1+\cos ^21+1\right)}\over{ 2\,\sin ^22+2\,\cos ^22+4\,\cos 2+2}}-{{\cos ^22\,\log \left(\sin ^2 1+2\,\sin 1+\cos ^21+1\right)}\over{2\,\sin ^22+2\,\cos ^22+4\,\cos 2+2}}-{{\log \left(\sin ^21+2\,\sin 1+\cos ^21+1\right)}\over{2\, \sin ^22+2\,\cos ^22+4\,\cos 2+2}}-{{\cos 2\,\log \left(\sin ^21+2\, \sin 1+\cos ^21+1\right)}\over{\sin ^22+\cos ^22+2\,\cos 2+1}}+{{ \sin ^22\,\log \left(\sin ^21-2\,\sin 1+\cos ^21+1\right)}\over{2\, \sin ^22+2\,\cos ^22+4\,\cos 2+2}}+{{\cos ^22\,\log \left(\sin ^21-2 \,\sin 1+\cos ^21+1\right)}\over{2\,\sin ^22+2\,\cos ^22+4\,\cos 2+2 }}+{{\log \left(\sin ^21-2\,\sin 1+\cos ^21+1\right)}\over{2\,\sin ^ 22+2\,\cos ^22+4\,\cos 2+2}}+{{\cos 2\,\log \left(\sin ^21-2\,\sin 1 +\cos ^21+1\right)}\over{\sin ^22+\cos ^22+2\,\cos 2+1}}+{{2\,\sin 1 \,\sin 2}\over{\sin ^22+\cos ^22+2\,\cos 2+1}}+{{2\,\cos 1\,\cos 2 }\over{\sin ^22+\cos ^22+2\,\cos 2+1}}+{{2\,\cos 1}\over{\sin ^22+ \cos ^22+2\,\cos 2+1}}$$

Численный ответ

0.624624546797409

Ответ (Неопределённый)

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 | x*sec(x)*tan(x) dx = C +  | x*sec(x)*tan(x) dx
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$$-{{\left(\sin ^2\left(2\,x\right)+\cos ^2\left(2\,x\right)+2\,\cos \left(2\,x\right)+1\right)\,\log \left(\sin ^2x+2\,\sin x+\cos ^2x+1 \right)+\left(-\sin ^2\left(2\,x\right)-\cos ^2\left(2\,x\right)-2\, \cos \left(2\,x\right)-1\right)\,\log \left(\sin ^2x-2\,\sin x+\cos ^2x+1\right)-4\,x\,\sin x\,\sin \left(2\,x\right)-4\,x\,\cos x\, \cos \left(2\,x\right)-4\,x\,\cos x}\over{2\,\sin ^2\left(2\,x \right)+2\,\cos ^2\left(2\,x\right)+4\,\cos \left(2\,x\right)+2}}$$