Pure Mathematics
posted by KerryAnn .
In=∫ x^nSinx dx....with the limits π/2 and 0
i found that In would be equal to n(π/2)^(n1) n(n1) In2
how do i deduce the value of I4? where n=4

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).
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