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`        1+cos4x/cotx-tanx`
8 years ago

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
```
8 years ago
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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```
8 years ago
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