# 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+...

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.

giorgiana1976 | Student

(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???

giorgiana1976 | Student

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)