Question

Help Trigonometric Substitution
\[\int\limits_{}^{}\sqrt{64-x ^{2}}\]

Answer 1

@hartnn

Answer 2

\[\cos^2 \theta = 1 - \sin^2 \theta\]

Answer 3

ok

Answer 4

Do you see how that helps us here?

Answer 5

yes because that will get you \[\sqrt{64\cos ^{2}\theta }\]

Answer 6

and then dx=8cos t dt
t for theta

Answer 7

Good. Keep going!

Answer 8

now this is where i'm getting kind of lost

Answer 9

The expression will be kind of messy, I think. You'll have to integrate it by parts. Do you know how to do that?

Answer 10

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

Answer 11

and then i'll work on integrating that

Answer 12

Sure, one sec

Answer 13

Actually I think I spaced out a bit.
I'm just getting \(\int d\theta\)

Answer 14

no the answer is something weird
\[\frac{ x }{ 2 }\sqrt{64-x ^{2}}+32\sin^{-1} (\frac{ x }{ 8 })+C\]

Answer 15

Ok, that looks like the result of integration by parts after all.
So how did you get that? :D

Answer 16

back of the book lol

Answer 17

that's why I'm having trouble I can't get there

Answer 18

So what did you @MATTW20 let x=?

Answer 19

Oh I see you let x=8sin(theta)
and then you said dx=8cos(theta) d theta
Just plug into your expression.

Answer 20

\[\int\limits_{}^{}\sqrt{64\cos^2(\theta) } 8 \cos(\theta) d \theta \]
That is what you have got so far, right?

Answer 21

correct

Answer 22

\[\int\limits_{}^{}8 \cos(\theta) \cos(\theta) d \theta ,\text{ assuming } \cos(\theta)>0\]
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)

Answer 23

\[put x=8\sin \theta ,dx=8\cos \theta d \theta \]
\[I=\int\limits \sqrt{64-64\sin ^{2}\theta} 8\cos \theta d \theta =32\int\limits 2\cos ^{2}\theta d \theta\]
\[=32\int\limits \left( 1+\cos 2\theta \right)d \theta=32\left( \theta+\frac{ \sin 2\theta }{2 } \right)+c\]

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

Answer 24

\[\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)=\cos^2(\theta)-(1-\cos^2(\theta))\]
\[\cos(\theta)=\cos^2(\theta)-1+\cos^2(\theta)=2\cos^2(\theta)-1\]
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.
\[\frac{1}{2}(\cos(\theta)+1)=\cos^2(\theta) \]

Answer 25

ok

Answer 26

And yes I forgot to write the other 8.. Errr...

Answer 27

\[\Large \color{teal}{\cos^2(\theta)\quad=\quad \frac{1}{2}(\cos2\theta+1)}\]Woops! Careful with your half-angle formula there missy! c:

Answer 28

Yeah I missed my 2.