"heavy vgs integral"

[Janhaa]

[tex]\\I=\int_{0}^{\pi/2}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\,dx[/tex]

[Kay offline]

[tex]\\I=\int_{0}^{\pi/2}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\,dx[/tex]

[+] Skjult tekst

Her anvender vi trikset [tex]\int_a^b f(x)dx=\int_a^bf(a+b-x)dx[/tex] (forutsatt at [tex]n[/tex] er et naturlig tall).

Så vi får at [tex]\int_{0}^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}dx=\int_0^{\frac{\pi}{2}}\frac{\sin^{n}\left ( \frac{\pi}{2}-x \right )}{\sin^n\left ( \frac{\pi}{2}-x \right )+\cos^n\left ( \frac{\pi}{2}-x \right )}dx=\int_0^\frac{\pi}{2} \frac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}dx[/tex]

Ved å definere det opprinnelige integralet som [tex]I[/tex] kan vi lett addere de sammen og se at [tex]2I=\int_0^\frac{\pi}{2}dx \Rightarrow I = \frac{\pi}{4}[/tex]

[Janhaa]

[tex]\\I=\int_{0}^{\pi/2}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}\,dx[/tex]

[+] Skjult tekst

Her anvender vi trikset [tex]\int_a^b f(x)dx=\int_a^bf(a+b-x)dx[/tex] (forutsatt at [tex]n[/tex] er et naturlig tall).

Så vi får at [tex]\int_{0}^{\frac{\pi}{2}}\frac{\sin^n(x)}{\sin^n(x)+\cos^n(x)}dx=\int_0^{\frac{\pi}{2}}\frac{\sin^{n}\left ( \frac{\pi}{2}-x \right )}{\sin^n\left ( \frac{\pi}{2}-x \right )+\cos^n\left ( \frac{\pi}{2}-x \right )}dx=\int_0^\frac{\pi}{2} \frac{\cos^n(x)}{\cos^n(x)+\sin^n(x)}dx[/tex]

Ved å definere det opprinnelige integralet som [tex]I[/tex] kan vi lett addere de sammen og se at [tex]2I=\int_0^\frac{\pi}{2}dx \Rightarrow I = \frac{\pi}{4}[/tex]

pent