## MATTW20 Group Title Help Trigonometric Substitution $\int\limits_{}^{}\sqrt{64-x ^{2}}$ 9 months ago 9 months ago

1. MATTW20 Group Title

@hartnn

2. SACAPUNTAS Group Title

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

3. MATTW20 Group Title

ok

4. SACAPUNTAS Group Title

Do you see how that helps us here?

5. MATTW20 Group Title

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

6. MATTW20 Group Title

and then dx=8cos t dt t for theta

7. SACAPUNTAS Group Title

Good. Keep going!

8. MATTW20 Group Title

now this is where i'm getting kind of lost

9. SACAPUNTAS Group Title

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 Group Title

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 Group Title

and then i'll work on integrating that

12. SACAPUNTAS Group Title

Sure, one sec

13. SACAPUNTAS Group Title

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

14. MATTW20 Group Title

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

15. SACAPUNTAS Group Title

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

16. MATTW20 Group Title

back of the book lol

17. MATTW20 Group Title

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

18. myininaya Group Title

So what did you @MATTW20 let x=?

19. myininaya Group Title

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

20. myininaya Group Title

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

21. MATTW20 Group Title

correct

22. myininaya Group Title

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

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

$\cos(2\theta)=\cos^2(\theta)-\sin^2(\theta)=\cos^2(\theta)-(1-\cos^2(\theta))$ $\cos(\theta)=\cos^2(\theta)-1+\cos^2(\theta)=2\cos^2(\theta)-1$ 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 Group Title

ok

26. myininaya Group Title

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

27. zepdrix Group Title

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

28. myininaya Group Title

Yeah I missed my 2.