Question

Integral (sin(2x))/(1+cos(2x)) dx?

Answer 1

Let u= (1+ cos^2 (x) )

Then du/dx = d/dx (1+ cos^2 (x) )

                   = d/dx (1) + d/dx (cos^2 (x))

                   = 0 + d/dx (cos^2 (x))

                   = 2 cos x d/dx (cos (x))            

                   = 2 cos x sin x                                        

 => du = 2 cos x sin x dx

              = sin 2x dx                       [ Since sin(2x) = 2sin x cos x ]

Therefore ∫ (sin(2x)) / (1+ cos^2 (x)) = ∫ 1/u du

                                                         = ln(u) +C                            [ Since ∫ 1/u du = ln(u) +C ]

                                                         = ln( 1+ cos^2(x) ) +C          [ Since u= (1+ cos^2 (x) ) ]

Therefore ∫ (sin(2x)) / (1+ cos^2 (x)) = ln( 1+ cos^2(x) ) +C

Answer 2

The function is ʃ { sin(2x)/[1+cos(2x)] } dx.

Let [ 1 + cos(2x) ] = u

- 2 sin(2x) dx = du

sin(2x) dx = - du/2

ʃ { sin(2x)/[1+cos(2x)] } dx = ʃ (- du/2)/u

                                      = - 1/2 ʃ (1/u) du

                                      = - 1/2 log (u)

                                      = - log √(u)

                                      = - log √[ 1 + cos(2x) ]

                                      = - log √[ 1 + {2cos²(x) - 1} ]

                                      = - log √[ 2cos²(x) ]

                                      = - log [√2cos (x)] + c

ʃ { sin(2x)/[1+cos(2x)] } dx = - log [√2 cos (x)] + c.

Comments

ʃ { sin(2x)/[1+cos(2x)] } dx = - log [√2 cos (x)] + C

= - log [ √2 ] - log [ cos (x)] + C

= - log [ cos (x)] + C - log [ √2 ]

= - log [ cos (x)] + C             (Since, log [√ 2 ] = 0.15 = constant)

 ∴ ʃ { sin(2x)/[1+cos(2x)] } dx = - log [ cos (x)] + C.