Recent content by CDevo69

Didn't know you could just move over the x. Sweet. Thankyou both quasar987 and courtrigrad

Some how the answer comes out to be: 1/2 (x^2sinx^2 + cosx^2) + c and I hope I didn't make a mistake in my calculations up there...

Well let's try... Int x^3cos(x^2) u = cos(x^2) du = -2x sin (x^2) dv = x^3 v = (x^4)/4 Int x^3cos(x^2) = (x^4cos(x^2))/4 + 1/2 Int x^5 sin (x^2) u = sin(x^2) du = 2x cos(x^2) dv = x^5 v = (x^6)/6 Int x^5 sin (x^2) = (x^6sin(x^2))/6 - 1/3 Int x^7 cos (x^2) Hmmm... Doesn't...

This has been posted before although I've come across it and got a different answer from https://www.physicsforums.com/archive/index.php/t-108378.html 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...

My school which is in Australia are doing it now. Its fairly easy to learn.