There are two ways to solve this.

**1.** `int2sinxcosx dx=2intsinxcosxdx=|[t=sinx],[dt=cosx dx]|=2inttdt=`

`(2t^2)/2 + C=sin^2x+C, C in RR`

**2.** Using formula for sine of double angle: `sin(2x)=2sinxcosx` so it follows:

`int2sinxcosx dx=int sin2xdx=|[t=2x],[dt=2dx=>dx=dt/2]|=`

`=1/2int sint dt=1/2(-cos t) +C = -1/2 cos 2x +C, C in RR`

These solutions may seam different but they are actually equal because `sin^2x=1/2-1/2cos2x` and since `1/2inRR` we may think of as of our constant `C`.