How do you find the integral of #(Sin2x) / (1 + cos^2x) dx#?

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Answer 1 · 2018-03-18T14:37:33

#=> -ln|cos2x + 3 | + c #

Use trig identities...

#cos2x = cos^2 x - sin^2 x = 2cos^2 x -1 #

#=> ( cos2x +1 )/2 +1 = cos^2 x + 1#

#=> ( cos2x + 3 )/2 = cos^2 x + 1 #

#=> int (sin2x) / (( cos2x +3 )/2) dx #

#=> int (2sin2x )/ (cos2x +3 ) dx #

Let #u = cos2x +3 #

#du = -2sin2x dx#

#=> -du = 2sin2x dx #

#=> -int (du)/u #

#=> -ln|u| + c #

#=> -ln|cos2x + 3 | + c #

#c - "constant" #

Answer 2 · 2018-03-18T14:40:11

Second method

#u = 1 + cos^2 x #

#=> du = 2cosx * -sinx dx #

#=> -du = 2sinxcosx dx #

#=> -du = sin2x dx #

#=> -int (du)/u #

#=> -ln|u| + c #

#=> -ln|1+cos^2 x | + c #