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

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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.
 
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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

\int_0^{\frac{2}{3}} 2x^3 \cos \theta \, dx

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

Also, how come when you made the trig substitution for x^3, you got

\sin^3 \theta

instead of

\left( \frac{4}{9} \right)^{\frac{3}{2}} \sin^3 \theta

?
 
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

\int_0^{\frac{2}{3}} 2x^3 \cos \theta \, dx

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

Also, how come when you made the trig substitution for x^3, you got

\sin^3 \theta

instead of

\left( \frac{4}{9} \right)^{\frac{3}{2}} \sin^3 \theta

?[/QUOTE
Oops, sorry, that was a stupid mistake on my part (leaving out sqrt(4/3))
 
Never mind him, was my answer right?(.133333333)
 
The answer is 64/1215.
 
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: \int_0 ^ \frac{2}{3} \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx. 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:

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

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:
\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)

Now, hopefully, you can take it from here, right? :)
 
  • #10
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
 
  • #11
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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  • #12
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:
 
  • #13
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. :) ^.^
 
  • #14
32cos^3 theta/243+32cos^5 theta/405
 
  • #15
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 :)
 
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