x sinx cosx = 1/2 x sin2x
u = x/2
du = 1/2 dx
dv = sin2x dx
v = -1/2 cos2x
∫u dv = uv - v du
= -1/4 x cos2x + 1/4 ∫cos2x dx
= -1/4 x cos2x + 1/8 sin2x + C
using the limits of integration, we have
(-1/4 π + 0) - (0 +0)
= -π/4
Solve the integral from [0,pi] of x*sinx*cosx*dx
I don't understand how to solve this problem because if I let u =x and du =1 and dv=sinx*cosx, I can't find v.
How do you solve this?
2 answers
I appreciate your kindness! Calc II is really a hard subject for me.