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

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evilpostingmong
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Homework Statement



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

Homework Equations





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!
 
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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.
 
Last edited:
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]

?
 
cepheid said:
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))
 
Never mind him, was my answer right?(.133333333)
 
Omg, no LaTeX. My eyes are hurting. :cry: :cry: :cry:
You should try to learn LaTeX, it's extremely convenient.
evilpostingmong said:

Homework Statement



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

You mean this: [tex]\int_0 ^ \frac{2}{3} \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx[/tex]. Right?

Homework Equations


The Attempt at a Solution


x=sqrt(4/9)sin theta
dx=sqrt(4/9)cosine theta

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)

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

So far, so good. :)

Integralx^3*2cos theta dtheta
Integral sin^3*2cos theta dtheta
Integral (cos^2theta-1)*sin theta*2cos theta dtheta

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.

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!

Homework Statement


Homework Equations


The Attempt at a Solution


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? :)
 
Wow, thanks VietDao, but I'm going to 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
 
evilpostingmong said:
Wow, thanks VietDao, but I'm going to 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

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? :)
 
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
 
evilpostingmong said:
Integral 32/81 (1-cos^2theta)*cos theta*sin theta d theta

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. :)

u=cos theta
du=sin theta

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

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

No... It's still wrong. :cry: :cry: :cry: :cry:
-(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 :)