Integral regning
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Hvordan fikk du den til å bli tanx? eller takk for hjelpen.sEirik wrote:Husk dx
[tex]I = \int \frac{{\rm d}x}{\cos^2 (2x)}[/tex]
[tex]u = 2x[/tex]
[tex]I = \frac{1}{2} \int \frac{{\rm d}u}{\cos^2 (u)}[/tex]
[tex]I = \frac{1}{2}\tan (u) + C = \frac{1}{2}\tan (2x) + C[/tex]
Just Remember u have afriend, when tRoubles seem like never end...!!
ja, men blir den ikke tan[sup]2[/sup]x ??sEirik wrote:[tex]\int \frac{1}{\cos^2 (x)} {\rm d}x = \tan x + C[/tex]
En vanlig regel. Prøv å vise den ved å derivere tangens, som er sin/cos.
Just Remember u have afriend, when tRoubles seem like never end...!!
[tex]\tan^\prime (x) = \left ( \frac {\sin (x) }{\cos (x)} \right )^\prime = \frac{\sin^\prime (x)\cos(x) - sin(x)\cos^\prime (x)}{\cos^2 (x)}[/tex]
[tex] = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)}[/tex]
[tex]\tan^\prime (x) = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)} = \frac{1}{\cos^2(x)}[/tex]
Eventuelt:
[tex]\tan^\prime (x) = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)} = 1 + \frac{\sin^2 (x)}{\cos^2 (x)} = 1 + \tan^2 (x)[/tex]
[tex] = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)}[/tex]
[tex]\tan^\prime (x) = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)} = \frac{1}{\cos^2(x)}[/tex]
Eventuelt:
[tex]\tan^\prime (x) = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)} = 1 + \frac{\sin^2 (x)}{\cos^2 (x)} = 1 + \tan^2 (x)[/tex]
Okei nå skjønner jeg det...Tusen takksEirik wrote:[tex]\tan^\prime (x) = \left ( \frac {\sin (x) }{\cos (x)} \right )^\prime = \frac{\sin^\prime (x)\cos(x) - sin(x)\cos^\prime (x)}{\cos^2 (x)}[/tex]
[tex] = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)}[/tex]
[tex]\tan^\prime (x) = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)} = \frac{1}{\cos^2(x)}[/tex]
Eventuelt:
[tex]\tan^\prime (x) = \frac{\cos^2(x) + sin^2(x)}{\cos^2 (x)} = 1 + \frac{\sin^2 (x)}{\cos^2 (x)} = 1 + \tan^2 (x)[/tex]
Just Remember u have afriend, when tRoubles seem like never end...!!