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Another Part where my brain's Fried 
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#1
May705, 02:52 PM

P: 13

[tex] \int \sqrt{4x^2}^3 dx [/tex]
I'm thinking integrating by parts would work, with u being that root and dv being dx, but is that the right method and direction. I've tried it and it seems more complicated. 


#2
May705, 03:12 PM

P: 12

Use [tex]x = 2 \sin \theta [/tex]...
then, [tex]dx = 2 \cos \theta d\theta [/tex], and your integral becomes: [tex]16 \int \cos^4 \theta d\theta [/tex] . Use [tex]\cos{2\theta} = 2\cos^2 \theta  1 [/tex], etc... 


#3
May705, 03:13 PM

P: 551

Damn it, no wonder I couldn't get it to work (I tried x = 2cos u)!



#4
May705, 03:13 PM

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Another Part where my brain's Fried
So it's
[tex]\int \left(4x^{2}\right)^{\frac{3}{2}} \ dx[/tex] How about a substitution [tex] x=2\sin t [/tex] ? Daniel. EDIT:Didn't see the other posts. 


#5
May705, 03:23 PM

Math
Emeritus
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PF Gold
P: 39,565

x= 2cos(u) should work exactly like "x= 2 sin(u)": dx= 2 sin u du and
[tex]\sqrt{4 x^2}= \sqrt{44 cos^2(u)}= 2 sin(u)[/tex] so the integral becomes [tex]16\int sin^4(u) du[/tex] and the only difference is that "". 


#6
May705, 03:26 PM

P: 551




#7
May805, 11:18 AM

P: 13

Thanks once more! :)



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