Integralx^3sqrt(4-9x^2)dx with limits of integration at 2/3, 0 (trig subst)

evilpostingmong

1. The problem statement, all variables and given/known data

Integralx^3sqrt(4-9x^2)dx with limits of integration at 2/3, 0 (trig subst)

2. Relevant equations



3. The attempt at a solution
x=sqrt(4/9)sin theta
dx=sqrt(4/9)cosine theta

sqrt(4-9x^2)
sqrt(4-9*4/9sin^2theta)
sqrt(4-4sin^2theta)
sqrt(4(1-sin^2theta)
sqrt(4cos^2theta)
2costheta

Integralx^3*2cos theta dtheta
Integral sin^3*2cos theta dtheta
Integral (cos^2theta-1)*sin theta*2cos theta dtheta
Integral (2cos^3 theta*sin theta dtheta)-Integral (2cos theta sin theta dtheta)
u=cos theta
du=sin theta dtheta
2u^4/4du-2u^2/2du
(2cos^4 theta/4)-(2cos^2 theta/2)

since x=sqrt(4/9)sin theta
x=opposite
sqrt(4/9)=hypoteneuse
and sqrt(4-9x^2)=adjacent
so 1/2sqrt(4-9x^2)^4-2/2sqrt(4-9x^2)^2+C
then plug in the numbers.

The number that I got was different from the books. Help is appreciated
since this problem is driving me crazy. Thank you!
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution

GoldPheonix

Remember to keep your coefficients well marked up. =P

I haven't finished the problem yet, but I can already tell you did that wrong. Go back and do them long hand:

u = a*sin(theta)
du = a*cos(theta) d(theta)


EDIT:
The answer I got was .1333333 repeating. Was that right? If it is, then check over your beginning again for coefficient mistakes.

cepheid

Okay, let's go to the step where you've just carried out the trig substitution for x in one of the factors, but not the other. You have

[tex] \int_0^{\frac{2}{3}} 2x^3 \cos \theta \, dx [/tex]

My question is, how come you wrote [itex] d\theta [/itex] instead of [itex] dx [/itex], without first converting the [itex] dx [/itex] into [itex] d\theta [/itex] properly?

Also, how come when you made the trig substitution for [itex] x^3 [/itex], you got

[tex] \sin^3 \theta [/tex]

instead of

[tex] \left( \frac{4}{9} \right)^{\frac{3}{2}} \sin^3 \theta [/tex]

?

evilpostingmong

[QUOTE=cepheid;1382003]Okay, let's go to the step where you've just carried out the trig substitution for x in one of the factors, but not the other. You have

[tex] \int_0^{\frac{2}{3}} 2x^3 \cos \theta \, dx [/tex]

My question is, how come you wrote [itex] d\theta [/itex] instead of [itex] dx [/itex], without first converting the [itex] dx [/itex] into [itex] d\theta [/itex] properly?

Also, how come when you made the trig substitution for [itex] x^3 [/itex], you got

[tex] \sin^3 \theta [/tex]

instead of

[tex] \left( \frac{4}{9} \right)^{\frac{3}{2}} \sin^3 \theta [/tex]

?[/QUOTE
Oops, sorry, that was a stupid mistake on my part (leaving out sqrt(4/3))

GoldPheonix

Never mind him, was my answer right?


(.133333333)

evilpostingmong

The answer is 64/1215.

VietDao29

Omg, no LaTeX. My eyes are hurting.
You should try to learn LaTeX, it's extremely convenient.
You mean this: [tex]\int_0 ^ \frac{2}{3} \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx[/tex]. Right?

Well, sqrt(4/9) is actually 2/3. You can rewritten the whole thing as:
x = (2/3) sin theta
dx = (2/3) cos theta d(theta)

So far, so good. :)

This is where you went wrong.
When chaging dx to d(theta), and x to sin theta, and cos theta you forget the factors.

x = (2/3) sin theta
dx = (2/3) cos theta d(theta)

You accidentally dropped out the 2/3 factor.

Well, you should re-do this part. :)

-----------------------------------

There's one more convenient way to tackle this problem, i.e to use u-substitution. x³, can be splitted into x² x

Letting u = x²

The whole thing become:

[tex]\int \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx = \frac{1}{2} \int \ u \sqrt{4 - 9u} \ du[/tex]

Another u-substitution will do it. Can you go from here? :)

evilpostingmong

Wow, thanks VietDao, but I'm gonna see if I could get it done the traditional way to see where I went wrong barring the stupid mistake..........

Integralx^3*2cos theta dtheta
Integral (2/3)sin^3*2cos theta dtheta
Integral 2/3(cos^2theta-1)*sin theta*2cos theta dtheta
Integral 4/3cos^3theta*sin theta d theta-Integral 4/3 cos theta sin theta d theta
u=cos theta
du=sin theta d theta

4u^4/12-4u^2/6
4cos^4 theta/12-4cos^2 theta/6+C

VietDao29

Whoops, it's still wrong =.=" You've messed up the coefficients.

Ok, let's do it step by step then.
Since x = (2/3) sin(theta)
x³ = (2/3)³ sin³(theta)
dx = (2/3) cos(theta) d(theta)

Change all x to theta, we have:
[tex]\int \left( \frac{2}{3} \sin ( \theta ) \right) ^ 3 \sqrt{4 - 9 \times \frac{4}{9} \sin ^ 2 \theta} \left( \frac{2}{3} \cos \theta \ d ( \theta ) \right) = \int \frac{8}{27} \sin ^ 3 ( \theta ) \left( 2 \cos \theta \right) \left( \frac{2}{3} \cos \theta \ d ( \theta ) \right)[/tex]

Now, hopefully, you can take it from here, right? :)

evilpostingmong

Integral 32/81 (1-cos^2theta)*cos theta*sin theta d theta
Integral 32/81 cos theta *sin theta d theta-Integral 32/81 cos^3 theta * sin theta d theta
u=cos theta
du=sin theta
32u^2/162-32u^4/324
32cos^2 theta/162-32cos^4 theta/324

VietDao29

Well, the first line is incorrect. >"<
Actually, it is cos²(theta) instead of cos(theta). Common, try again one more time, man, you are very very close to the answer. Just correct some minor mistakes, and it's done. :)

Oh, and by the way. If u = cos(theta), then du would be: du = -sin(theta)d(theta)

evilpostingmong

Then I'll just simply change the first line to
Integral 32/81 cos^2 theta sin theta d theta- Integral 32/81 cos^4 theta sin theta d theta
u=cos theta
du= -sin theta d theta
32u^3/243-32u^5/405
32 cos^3 theta/243-32 cos^5 theta/405

VietDao29

Well, you still got the wrong signs, though. du = -sin(theta)d(theta). There's a minus sign in front of it. :)

Ok, well, yeah, just change the signs, and you'll arrive at the final answer. :) ^.^

evilpostingmong

32cos^3 theta/243+32cos^5 theta/405

VietDao29

No... It's still wrong.
-(a - b) = -a + b, not a + b

You should flip both signs.
-32 cos^3 theta/243+32 cos^5 theta/405

Well, you should do some more manipulations, it's good for some problem like this, i.e, require a lot of calculation.

Btw, congratulations. Finally, you got it :)