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Grade: 12

                        

1+cos4x/cotx-tanx

10 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

10 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


Please feel free to post as many doubts on our discussion forum as you can. If you find any

question Difficult to understand - post it here and we will get you the answer and detailed

solution very quickly.
 We are all IITians and here to help you in your IIT JEE preparation.

 All the best.
 
Regards,
Askiitians Experts
Badiuddin

 

 

10 years ago
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