2 Answers chandra sekhar
10 Points
13 years ago

Hi Paresh,

I= ∫1+ cos4x*sinx* cosx/(cos2x-sin2x)  dx

  = ∫1+ (2cos22x-1)*sin2x/2cos2x  dx

let cos2x = t

I= ∫1+ (2t2-1)/(-4t) dt

 =∫(1-t/2+1/4t) dt

 =t - t2/4 + 1/4 ln t + c     where cos2x=t

Badiuddin askIITians.ismu Expert
148 Points
13 years ago

Hi Paresh,

I=(1+cos4x)/(cotx-tanx) dx

I= ∫(1+ cos4x)sinx cosx/(cos2x-sin2x)  dx

  = ∫(1+ 2cos22x-1)sin2x/2cos2x  dx

I= ∫cos2x sin2xdx

   = 1/2 ∫ sin4xdx

   = 1/8 [-cos4x] +c

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