#int(1+cos(2x))/(sin(2x))dx=?

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Answer 1 · 2018-04-14T15:07:22

The answer is #"option (B)"#

We need

#int(cotx)=ln|sinx|+C#

#intcscx=ln(|cscx+cotx|)+C#

Let #u=2x#, #=>#, #du=2dx#

The integral is

#I=int((1+cos2x)dx)/(sin2x)=1/2int((1+cosu)du)/(sinu)#

#=1/2int(cotu+cscu)du#

#=1/2(ln(sinu))-1/2ln(cscu+cotu)#

#=1/2ln(sinu)-1/2ln((1+cosu)/sinu)#

#=1/2ln(sin2x)-1/2ln(1+cos2x)-1/2ln(sin2x)+C#

#=-1/2ln(1+2cos^2 (2x)-1)#

#=-1/2*2lncos(2x)+C#

#=-ln(|cos2x|)+C#

The answer is #"option (B)"#