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Finn $\int_0^{\frac{\pi}{2}}\frac{1}{1+(\tan x)^{\sqrt{2}}}\, dx$

Generelt har man at $\int_a^b f(x)~\mathrm{d}x = \int_a^b f(a+b-x)~\mathrm{d}x$.
Dette gir $I = \int_0^{\pi/2}\dfrac{1}{1+(\tan{x})^\sqrt{2}}\mathrm{d}x = \int_0^{\pi/2}\dfrac{1}{1+(\tan{(\pi/2 - x)})^\sqrt{2}}\mathrm{d}x$
Her er $\tan(\pi/2-x) = \dfrac{\sin(\pi/2-x)}{\cos(\pi/2-x)} = \dfrac{\cos{x}}{\sin{x}} = \cot{x}$, slik at $I = \int_0^{\pi/2}\dfrac{1}{1+(\cot{x})^\sqrt{2}}\mathrm{d}x = \int_0^{\pi/2}\dfrac{(\tan(x))^\sqrt{2}}{1+(\tan(x))^\sqrt{2}}\mathrm{d}x$
og $2I = \int_0^{\pi/2}\dfrac{1}{1+(\tan{x})^\sqrt{2}}\mathrm{d}x+\int_0^{\pi/2}\dfrac{(\tan(x))^\sqrt{2}}{1+(\tan(x))^\sqrt{2}}\mathrm{d}x = \int_0^{\pi/2}\dfrac{1+(\tan(x))^\sqrt{2}}{1+(\tan(x))^\sqrt{2}}\mathrm{d}x = \pi/2\Longrightarrow I = \pi/4$