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Question

Solve sin2xacos2x+bsin2x

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Solution

Consider the given function:

sin2xacos2x+bsin2xdx

putacos2x+bsin2x=t

(a.2cosxsinx+b.2sinxcosx)dxdt=1

(ba).sin2xdx=dt

sin2xdx=1badt

sin2xacos2x+bsin2xdx

=1ba1tdt

=1balogt+c

=1balog(acos2x+bsin2x)+c

Hence this is the answer.

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