Đáp án:
a. Đặt u = x => du = dx
dv=cos2xsinxdx⇒v=cosx1dx⇒I=cosxx∣∣−3π3π−∫−3π3πcosxdxI=34π−∫−3π3πsin(x+2π)dx=34π−∫−3π3π2sin(2x+2π)cos(x+4π)dx=34π−lntan∣∣2x+4π∣∣∣∣−3π3π=34π−2ln(tan(125π))
b. Đặt u = x + sinx => du = (1 + cosx)dx
dv=1+cosxdx⇒v=tan2xJ=(x+sinx)tan2x∣∣02π−∫02π(1+cosx)2xdx=(x+sinx)tan2x∣∣02π−∫02πsinxdx=2π+2