Solving Surface Integral: x(C - x^2/3)^3/2 dx

[gulsen]

$$\int x {(C - x^{2/3})}^{3/2} dx$$

Any ideas?

 

[VietDao29]

$$\int x {(C - x^{2/3})}^{3/2} dx$$

Any ideas?

$$\int x \sqrt{ \left( C - \sqrt[3]{x ^ 2} \right) ^ 3} dx$$
To solve integrals with square roots like that, one can start off their problem by changing the variable x to some variable t such that:
$$C - \sqrt[3]{x ^ 2} = C \cos ^ 2 t$$
So, let:
$$x = \sqrt{C ^ 3} \sin ^ 3 t \quad \quad \quad t \in \left[ -\frac{\pi}{2} ; \ \frac{\pi}{2} \right]$$
$$\Rightarrow dx = 3 \sqrt{C ^ 3} \cos t \sin ^ 2 t dt$$.
Substitute all to your integral yields:
$$\int x \sqrt{\left( C - \sqrt[3]{x ^ 2} \right) ^ 3} dx = 3 C ^ 3 \int \sin ^ 5 t \cos t \sqrt{ \left( C - \sqrt[3]{C ^ 3 \sin ^ 6 t} \right) ^ 3} dt = 3 C ^ 3 \int \sin ^ 5 t \cos t \sqrt{ \left( C - C \sin ^ 2 t} \right) ^ 3} dt$$
$$= 3 \sqrt{C ^ 9} \int \sin ^ 5 t \cos ^ 4 t dt$$
You can go from here, right? :)

 

[gulsen]

Yup, thanks! :)
But isn't this substitution very subtle? Am I an oracle to guess it instantenously in an exam!?

 

[masudr]

Well it may appear ingenious, but really $sin^2t + cos^2t=1$ so the idea of substituting some function of $sin t$ is going to be a good idea. Once you have that idea, you can work out what powers you need &c.

 

[VietDao29]

Yup, thanks! :)
But isn't this substitution very subtle? Am I an oracle to guess it instantenously in an exam!?

Nah, you don't need to be an oracle to know it.
To solve square root, there are some common ways:
The first way is to use an u substitution:
u² = everything inside the square root, so that when you squre root it, it'll become u. In your problem:
$$u ^ 2 = C - \sqrt[3]{x ^ 2}$$.
Then $$2u du = -\frac{2}{3 \sqrt[3]{x}} dx \Rightarrow dx = 3u \sqrt[3]{x}$$.
Substitute all to your integral, and it will become:
$$\int x \sqrt{ \left( C - \sqrt[3]{x ^ 2} \right) ^ 3} dx = 3 \int u ^ 4 \sqrt[3]{x ^ 4} du = 3 \int u ^ 4 (C - u ^ 2) ^ 2 du$$
Try it to see if you can arrive at the same answer as you previously got.
------------------
The second way is to use trig substitution. Change x to u such that everything inside the square root become something * cos²u, so that again when you square root it, it will be come: $$\sqrt{\mbox{something}} \times \cos u$$.
So let:
$$C - \sqrt[3]{x ^ 2} = C \cos ^ 2 t$$
$$\Leftrightarrow \sqrt[3]{x ^ 2} = C \sin ^ 2 t$$
$$\Leftrightarrow x = \sqrt{C ^ 3} \sin ^ 3 t$$
$$\Rightarrow dx = 3 \sqrt{C ^ 3} \cos t \sin ^ 2 t dt$$
...
Can you get this? :)

 

[arildno]

Alternatively, you may choose:
$$y=x^{\frac{1}{3}}\to{dx}=3y^{2}dy$$
And, hence:
$$\int{x}(C-x^{\frac{2}{3}})^{\frac{3}{2}}dx=\int{3y^{5}}(C-y^{2})^{\frac{3}{2}}dy$$

This is solved nicely by repeated integrations by parts.

 

[daveyp225]

even better...

Let $$u = \sqrt{C-x^{\frac{2}{3}}$$


Then $$x = (C-u^2)^{\frac{3}{2}} \Rightarrow$$
...

$$dx = \frac{3}{2}(C-u^2)^{\frac{1}{2}}(-2)udu\Rightarrow$$

...

$$dx = -3(C-u^2)^{\frac{1}{2}}udu$$
...
Then: $$\int x(C-x^{\frac{2}{3}})^{\frac{3}{2}}dx = \int[ (C-u^2)^{\frac{3}{2}})(u^3)(-3(C-u^2)^{\frac{1}{2}}udu)]$$

Put the two $$(C-u^2)$$ terms together and simplifying:

$$= -3\int (C-u^2)^2u^4du \Rightarrow -3\int(c^2 - 2Cu^2 + u^4)(u^4)du$$

...
$$=-3\int(u^4c^2 - 2u^6C + u^8)du$$
...
$$=-3[C^2\int u^4du - 2C\int u^6 du + \int u^8 du]$$

:)

 

[VietDao29]

even better...
...

Have you looked closely at the fifth post of this thread, daveyp225??

 

[daveyp225]

Have you looked closely at the fifth post of this thread, daveyp225??

Evidently not.