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

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

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

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

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

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

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

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

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

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

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

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

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

myininaya Group TitleBest ResponseYou've already chosen the best response.1
So what did you @MATTW20 let x=?
 11 months 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.
 11 months 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?
 11 months 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)
 11 months 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 \[I=32\left( \theta+\frac{ 2\sin \theta \cos \theta }{ 2 } \right)+c\] \[I=32\left( \sin^{1} \frac{ x }{8 }+\frac{ x }{8 }*\cos \frac{ \sqrt{64x ^{2}} }{8 } \right)+c\]
 11 months 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. \[\frac{1}{2}(\cos(\theta)+1)=\cos^2(\theta) \]
 11 months ago

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

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