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

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
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

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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.
 Mentor 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 $$\left( \frac{4}{9} \right)^{\frac{3}{2}} \sin^3 \theta$$ ?

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

[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

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

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

$$\sin^3 \theta$$

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

?[/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.

Recognitions:
Homework Help
Omg, no LaTeX. My eyes are hurting.
You should try to learn LaTeX, it's extremely convenient.
 Quote by 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)
You mean this: $$\int_0 ^ \frac{2}{3} \ x ^ 3 \sqrt{4 - 9 x ^ 2} \ dx$$. Right?

 2. Relevant equations 3. 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! 1. The problem statement, all variables and given/known data 2. Relevant equations 3. 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 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

Recognitions:
Homework Help
 Quote by 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
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? :)

 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

Recognitions:
Homework Help
 Quote by evilpostingmong 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)

 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

Recognitions:
Homework Help
 Quote by 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
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. :) ^.^

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

Recognitions:
Homework Help
 Quote by evilpostingmong 32cos^3 theta/243+32cos^5 theta/405
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 :)