The correct option is C
−cos 2x−log(sec x|+C
We know the reduction formula for In=∫sin nxcos xdx as :
Now, to find In=∫sin 3xcos xdx
we take n=3, thus substituting it back in the reduction formula we get
Thus we get I3=−cos 2x−ln(secx|+C
So, Option c. is correct.