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

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

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

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

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

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

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

MATTW20
 one year ago
Best ResponseYou've already chosen the best response.0Yeah 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

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

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

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

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

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

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

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

myininaya
 one year ago
Best 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?

myininaya
 one year ago
Best 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)

surjithayer
 one year ago
Best 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 \[I=32\left( \theta+\frac{ 2\sin \theta \cos \theta }{ 2 } \right)+c\] \[I=32\left( \sin^{1} \frac{ x }{8 }+\frac{ x }{8 }*\cos \frac{ \sqrt{64x ^{2}} }{8 } \right)+c\]

myininaya
 one year ago
Best 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. \[\frac{1}{2}(\cos(\theta)+1)=\cos^2(\theta) \]

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

zepdrix
 one year ago
Best 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:
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