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) *...

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)