If ∫cos3x2sin2xdx=(tanx)A+C(tanx)B+k, where k is a constant of integration, then A+B+C equals:
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Official solution
∫cos3x4sinxcosxdx=∫2cos4xtanxdx
Let tanx=t2⇒sec2xdx=2tdt, and sec2x=1+t4=∫2tanxsec4xdx=∫2tanxsec2x(sec2xdx)=∫2t(1+t4)⋅2tdt=∫(1+t4)dt=t+5t5+k=tanx+51tan5/2x+kA=21,B=25,C=51A+B+C=516