∫cos3x4sinxcosxdx=∫2cos4xtanxdx
Let, tanx=t2⇒sec2x=1+t4
And sec2xdx=2tdt
∴I=∫2tanxsec4xdx=∫2tanxsec2x.sec2xdx
=∫2t(1+t4)2tdt=∫(1+t4)dt
=t+5t5+k, where k is a constant of integration.
=tanx+51tan25x+k
Where, t=tanx.
A=21,B=25,C=51
A+B+C=516