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1+cos4x/cotx-tanx
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
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
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
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
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