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MATTW20
Group Title
Help Trigonometric Substitution
\[\int\limits_{}^{}\sqrt{64x ^{2}}\]
 one year ago
 one year ago
MATTW20 Group Title
Help Trigonometric Substitution \[\int\limits_{}^{}\sqrt{64x ^{2}}\]
 one year ago
 one year ago

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SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
\[\cos^2 \theta = 1  \sin^2 \theta\]
 one year ago

SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
Do you see how that helps us here?
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
yes because that will get you \[\sqrt{64\cos ^{2}\theta }\]
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
and then dx=8cos t dt t for theta
 one year ago

SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
Good. Keep going!
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
now this is where i'm getting kind of lost
 one year ago

SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
The expression will be kind of messy, I think. You'll have to integrate it by parts. Do you know how to do that?
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
Yeah we passed that section. I just am not super familiar with it. but could you just write the expression I'm supposed to integrate by parts so I make sure everything is correct up to that point
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
and then i'll work on integrating that
 one year ago

SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
Sure, one sec
 one year ago

SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
Actually I think I spaced out a bit. I'm just getting \(\int d\theta\)
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
no the answer is something weird \[\frac{ x }{ 2 }\sqrt{64x ^{2}}+32\sin^{1} (\frac{ x }{ 8 })+C\]
 one year ago

SACAPUNTAS Group TitleBest ResponseYou've already chosen the best response.1
Ok, that looks like the result of integration by parts after all. So how did you get that? :D
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
back of the book lol
 one year ago

MATTW20 Group TitleBest ResponseYou've already chosen the best response.0
that's why I'm having trouble I can't get there
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So what did you @MATTW20 let x=?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Oh I see you let x=8sin(theta) and then you said dx=8cos(theta) d theta Just plug into your expression.
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{}^{}\sqrt{64\cos^2(\theta) } 8 \cos(\theta) d \theta \] That is what you have got so far, right?
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\int\limits_{}^{}8 \cos(\theta) \cos(\theta) d \theta ,\text{ assuming } \cos(\theta)>0\] so we have that we are trying to evaluate: \[8\int\limits_{}^{}\cos^2(\theta) d \theta \] The double angle identity comes in use for integating cos^2(theta) and even sin^2(theta)
 one year ago

surjithayer Group TitleBest ResponseYou've already chosen the best response.1
\[put x=8\sin \theta ,dx=8\cos \theta d \theta \] \[I=\int\limits \sqrt{6464\sin ^{2}\theta} 8\cos \theta d \theta =32\int\limits 2\cos ^{2}\theta d \theta\] \[=32\int\limits \left( 1+\cos 2\theta \right)d \theta=32\left( \theta+\frac{ \sin 2\theta }{2 } \right)+c\] dw:1381785998999:dw \[I=32\left( \theta+\frac{ 2\sin \theta \cos \theta }{ 2 } \right)+c\] \[I=32\left( \sin^{1} \frac{ x }{8 }+\frac{ x }{8 }*\cos \frac{ \sqrt{64x ^{2}} }{8 } \right)+c\]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
\[\cos(2\theta)=\cos^2(\theta)\sin^2(\theta)=\cos^2(\theta)(1\cos^2(\theta))\] \[\cos(\theta)=\cos^2(\theta)1+\cos^2(\theta)=2\cos^2(\theta)1\] Solve for cos^2(theta) This is just the way I remember this formula. I remember other formulas and derive other formulas from the ones I have memorized. \[\frac{1}{2}(\cos(\theta)+1)=\cos^2(\theta) \]
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
And yes I forgot to write the other 8.. Errr...
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\Large \color{teal}{\cos^2(\theta)\quad=\quad \frac{1}{2}(\cos2\theta+1)}\]Woops! Careful with your halfangle formula there missy! c:
 one year ago

myininaya Group TitleBest ResponseYou've already chosen the best response.1
Yeah I missed my 2.
 one year ago
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