Posted by **Kerry-Ann** on Wednesday, January 15, 2014 at 10:51pm.

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

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

- Pure Mathematics -
**Steve**, Thursday, January 16, 2014 at 12:49am
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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