In=∫ x^nSinx dx....with the limits π/2 and 0

Question

i found that In would be equal to n(π/2)^(n-1) -n(n-1) In-2
how do i deduce the value of I4? where n=4

Answer 1

you need to do integration by parts.

∫x^4 sinx dx
u=x^4
du = 4x^3 dx

dv = sinx dx
v = -cosx

∫x^4 sinx = -x^4 cosx + 4∫x^3 cosx dx

Now do all that again for each resulting integral, and the powers of x descends until you are left with just ∫sinx dx

Then just add all the pieces together.

Hmmm. I think you can save yourself some work. At x=0 all the terms with powers of x and sinx vanish.

at x=π/2 all the terms with cosx vanish (odd powers of x).