#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 $"option (B)"$