## MATTW20 one year ago Help Trigonometric Substitution $\int\limits_{}^{}\sqrt{64-x ^{2}}$

1. MATTW20

@hartnn

2. SACAPUNTAS

$\cos^2 \theta = 1 - \sin^2 \theta$

3. MATTW20

ok

4. SACAPUNTAS

Do you see how that helps us here?

5. MATTW20

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

6. MATTW20

and then dx=8cos t dt t for theta

7. SACAPUNTAS

Good. Keep going!

8. MATTW20

now this is where i'm getting kind of lost

9. SACAPUNTAS

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

10. MATTW20

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. MATTW20

and then i'll work on integrating that

12. SACAPUNTAS

Sure, one sec

13. SACAPUNTAS

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

14. MATTW20

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

15. SACAPUNTAS

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

16. MATTW20

back of the book lol

17. MATTW20

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

18. myininaya

So what did you @MATTW20 let x=?

19. myininaya

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

20. myininaya

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

21. MATTW20

correct

22. myininaya

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

23. surjithayer

$put x=8\sin \theta ,dx=8\cos \theta d \theta$ $I=\int\limits \sqrt{64-64\sin ^{2}\theta} 8\cos \theta d \theta =32\int\limits 2\cos ^{2}\theta d \theta$ $=32\int\limits \left( 1+\cos 2\theta \right)d \theta=32\left( \theta+\frac{ \sin 2\theta }{2 } \right)+c$ |dw:1381785998999:dw| $I=32\left( \theta+\frac{ 2\sin \theta \cos \theta }{ 2 } \right)+c$ $I=32\left( \sin^{-1} \frac{ x }{8 }+\frac{ x }{8 }*\cos \frac{ \sqrt{64-x ^{2}} }{8 } \right)+c$

24. myininaya

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

25. MATTW20

ok

26. myininaya

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

27. zepdrix

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

28. myininaya

Yeah I missed my 2.