Calculer les intégrales suivantes:
\(\int_{0}^{2\pi} \cos^2 x\,\textrm{d}x\)
\(\int_{0}^{1} \dfrac{1}{\sqrt{1-x^2}}\,\textrm{d}x\)
\(\int_{1}^{\mathop{\mathrm{ch}}1} \dfrac{1}{\sqrt{x^2 -1}}\,\textrm{d}x\)
\(\int_{0}^{1} \dfrac{x}{1+x^4}\,\textrm{d}x\)
\(\int_{e}^{e^2} {\scriptstyle 1\over\scriptstyle x\left(\ln x\right)^2}\,\textrm{d}x\)
\(\int_{1+e}^{1+2e} \dfrac{x^2}{x-1}\,\textrm{d}x\)
\(\int_{0}^{2\pi} \cos^2 x\,\textrm{d}x= \int_{0}^{2\pi} {\scriptstyle 1+\cos 2x \over\scriptstyle 2}\,\textrm{d}x=\Bigl[ {\scriptstyle x\over\scriptstyle 2} - {\scriptstyle\sin 2x\over\scriptstyle 4} \Bigr]_{0}^{2\pi}=\boxed{\pi}\)
\(\int_{0}^{1} \dfrac{1}{\sqrt{1-x^2}}\,\textrm{d}x=\Bigl[ \operatorname{arcsin} x \Bigr]_{0}^{1}=\boxed{\dfrac{\pi}{2}}\)
\(\int_{0}^{\mathop{\mathrm{ch}}1} \dfrac{1}{\sqrt{x^2 -1}}\,\textrm{d}x=\Bigl[ \mathop{\mathrm{argch}}x \Bigr]_{1}^{\mathop{\mathrm{ch}} 1}=\boxed{1}\)
\(\int_{0}^{1} \dfrac{x}{1+x^4}\,\textrm{d}x=\Bigl[ \dfrac{\operatorname{arctan} x^2}{2} \Bigr]_{0}^{1}=\boxed{\dfrac{\pi}{8}}\)
\(\int_{e}^{e^2} {\scriptstyle 1\over\scriptstyle x\left(\ln x\right)^2}\,\textrm{d}x=\Bigl[ -\dfrac{1}{\ln x} \Bigr]_{e}^{e^2}=\boxed{\dfrac{1}{2}}\)
\(\int_{1+e}^{1+2e} \dfrac{x^2}{x-1}\,\textrm{d}x=\int_{0}^{1} \dfrac{x^2-1}{x-1}\,\textrm{d}x +\int_{0}^{1} \dfrac{1}{x-1}\,\textrm{d}x=\Bigl[ \dfrac{x^2}2+x+\ln\left|x-1\right| \Bigr]_{1+e}^{1+2e} =\boxed{2e+\dfrac{3}{2}e^2+\ln 2}\)