- #1

CDevo69

- 5

- 0

x sinx cosx dx using the identity sin2x = 2sinxcosx

u = x

du = 1

dv = 1/2 sin 2x

v = -1/4 cos 2x

x sinx cosx dx = -1/4xcos2x - Int -1/4cos2x + c

= -1/4xcos2x + 1/8sin2x + c

Right?

And I need some help to work out:

Integral of x^3cosx^2

I'm always getting confused with powers on the trig for some reason.

Thankyou