Evaluate ∫sin⁡xsin⁡3xdx

I=∫sin⁡xsin⁡3xdx=∫sin⁡x3sin⁡x−4sin3⁡xdx=∫13−4sin2⁡xdx=∫sec2⁡x3sec2⁡x−4tan2⁡xdx    Dividing Nr and Dr by cos2x     Putting tan x = t and sec2x dx = dt, we getI=∫dt31+t2−4t2=∫dt3−t2=∫1(3)2−t2dt=123log⁡3+t3−t+C=123log⁡3+tan⁡x3−tan⁡x+C