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paresh maheshwari Grade: 12
        

1+cos4x/cotx-tanx

7 years ago

Answers : (2)

askiitian.expert- chandra sekhar
10 Points
										

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

7 years ago
Badiuddin askIITians.ismu Expert
147 Points
										

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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7 years ago
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