
#19
Dec810, 03:44 PM

P: 2

I think I have a solution. I hope it was not so late :).
tan(x^2)=m dx=cos(x^2)dm integral [sin(x^2)] = integral [mdm/(m^2+1)] m^2+1=a and 2mdm=da ........ integral [sin(x^2)] = 0.5*ln[(tan(x^2))^2+1]+c 



#21
Dec810, 04:16 PM

PF Gold
P: 1,930





#22
Dec810, 04:40 PM

P: 2

Actualy is not true. I have done a misteke when calculete the [tan(x^2)]'
2xdx=cos^2(x^2)dm is true if I find a solution with this I ll write. 



#23
Dec810, 04:42 PM

PF Gold
P: 1,930

If tan(x^2) = m, then [tex]2x sec^2(x^2) dx = dm[/tex] and [tex]2x dx = cos^2(x^2) dm[/tex] Can you see why? 



#24
May812, 11:57 AM

P: 1

[tex]\left[cos(u)\right]_0^{625} = cos(625)cos(0) = (0.984387)(1)=1.984387 [/tex] Therefore; [tex]\frac{1}{4}\int_0^{625} sin(u) du = \frac{1}{4}(1.984387)= 0.496097[/tex] 


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