If ∫dxcos3x√2sin2x=(tanx)A+C(tanx)B+k, where k is a constant of integration, then the value of A+B+C is equal to:

I=∫dxcos3x√2sin2x
→I=∫sec3x2√sinxcosxdx
Multiplying numerator and and denominator of I by secx, we get
I=∫sec4x2√tanxdx
Now assuming, tanx=t² →sec²xdx=2tdt, we get
I=∫1+t⁴2t(2t) dt →I=∫(1+t⁴) dt →I=t+t⁵/5+K
Re-Substituting t=√tanx, we get
I=√tanx+1/5⋅(√tanx)⁵+K
Comparing it with the given expression
I=√tanx+1/5⋅(√tanx)⁵+K=(tanx)A+C(tanx)B+K
→A=1/2,B=5/2,C=1/5
A+B+C=1/2+5/2+1/5=16/5=165