I need to find the recurrence for integral from 0 to pi/2 sin^n(x) * cos(x) dx.
I made int 0-pi/2 sin^n(x) *cosx dx=int sin^n (x) * ((sinnx)/n)`..... and it gives me 1/n+ int(sin^n-1(x)*(cosnx*sinx-sinnx*cosx)dx, so i cant find the recurrence.
recurrence for int cos^n(x)*sinnx dx i managed to solve it.= I[sub]n[/sub]=I[sub]n-1[/sub]/2 +1/2n.
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(sin x)^n = sin ^n (x) are the same thing
omg, sorry for this mess but i cant find the edit button.
The integral its int 0 to pi/2 (sin^n)x * cosnx dx.
giorgiana1976, im sorry, but its (sin^n)(x). Its the same thing as (sinx)^n???
We write the integral as:
Int (sin x)^n*cos xdx
We notice that the derivative of sin x is cos x. We'll substitute:
sin x = t
We'll differentiate both sides:
cos xdx = dt
We'll re-write the integral in t:
Int t^n*dt = t^(n+1)/(n+1) + C
But t = sin x
Int (sin x)^n*cos xdx = (sin x)^(n+1)/(n+1) + C
We'll determine F(b) for b = pi/2:
F(pi/2) = (sin pi/2)^(n+1)/(n+1)
F(pi/2) = 1^(n+1)/(n+1)
F(pi/2) = 1/(n+1)
F(0) = 0, for sin 0 = 0
Int (sin x)^n*cos xdx = F(pi/2) - F(0)
Int (sin x)^n*cos xdx = 1/(n+1)
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