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anonymous
 3 years ago
Help Trigonometric Substitution
\[\int\limits_{}^{}\sqrt{64x ^{2}}\]
anonymous
 3 years ago
Help Trigonometric Substitution \[\int\limits_{}^{}\sqrt{64x ^{2}}\]

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\cos^2 \theta = 1  \sin^2 \theta\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Do you see how that helps us here?

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then dx=8cos t dt t for theta

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now this is where i'm getting kind of lost

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The expression will be kind of messy, I think. You'll have to integrate it by parts. Do you know how to do that?

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and then i'll work on integrating that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Actually I think I spaced out a bit. I'm just getting \(\int d\theta\)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, that looks like the result of integration by parts after all. So how did you get that? :D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that's why I'm having trouble I can't get there

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

myininaya
 3 years 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
 3 years 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
 3 years 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)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[put x=8\sin \theta ,dx=8\cos \theta d \theta \] \[I=\int\limits \sqrt{6464\sin ^{2}\theta} 8\cos \theta d \theta =32\int\limits 2\cos ^{2}\theta d \theta\] \[=32\int\limits \left( 1+\cos 2\theta \right)d \theta=32\left( \theta+\frac{ \sin 2\theta }{2 } \right)+c\] 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1And yes I forgot to write the other 8.. Errr...

zepdrix
 3 years 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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