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God trening i sub og trig!

[tex][tex][/tex]\int_{0}^{1}[ln(x+1)/(x^2+1]=(pi/8)*(ln(2))

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vilma skrev:God trening i sub og trig!

[tex][tex][/tex]\int_{0}^{1}[ln(x+1)/(x^2+1]=(pi/8)*(ln(2))

[tex]x=\tan(u)[/tex]

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vilma skrev:God trening i sub og trig!

[tex][tex][/tex]\int_{0}^{1}[ln(x+1)/(x^2+1]=(pi/8)*(ln(2))

[tex]x=\tan(u)[/tex]

[tex]dx=(1+\tan^2(u))du[/tex]

[tex]I=\int_{0}^{\pi/4}\ln(1+\tan(u))du\\ \\ \\bruker\\ \int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx\\I=\int_0^{\pi/4}\ln(1+\tan(\frac{\pi}{4}-u))\\ \tan(\frac{\pi}{4}-u)=\frac{1-\tan(u)}{1+\tan(u)}\\ \\[/tex]

[tex]I=\int_{0}^{\pi/4}\ln(1+\frac{1-\tan(u)}{1+\tan(u)})du=\int_{0}^{\pi/4}\ln(\frac{2}{1+\tan(u)})du\\ \\ I=\int_0^{\pi/4}\ln(2)du\,-\,\int_{0}^{\pi/4}\ln(1+\tan(u))du\\ \\2I=\frac{\pi}{4}\ln(2)\\ \\I=\frac{\pi}{8}\ln(2)[/tex]