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

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1 Answer
Apr 14, 2018

The answer is "option (B)"

Explanation:

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)"