Another trig sub problem

1. Jun 9, 2006

suspenc3

heres what i have:

$$\int_{0}^{2\sqrt{3}} \frac {x^3}{\sqrt{16-x^2}}dx$$
let $$x=4sin\phi$$

$$dx=4cos4\phi$$

$$\sqrt{16-x^2} = \sqrt{16-4sin^2\phi} = \sqrt{4cos^2\phi = 4cos\phi$$

$$\int_{0}^{2\sqrt{3}} \frac{x^3}{\sqrt{16-x^2}}dx$$

$$\int_{0}^{2\sqrt{3}} \frac{4sin^3\phi}{4cos\phi}4cos\phi d\phi$$

$$= 4\int_{0}^{2\sqrt{3}} sin^3/phi$$

$$=4\int_{0}^{2\sqrt{3}}sin^2\phi(sin(phi)$$

$$=4\int_{0}^{2\sqrt{3}} 1-cos^2\phi(sin\phi)$$

$$=4\int_{0}^{2\sqrt{3}} sin\phi - sin \phi (cos^2\phi)$$

this is as far as i went because i think i messed up

any help would be appreciated

Last edited: Jun 9, 2006
2. Jun 9, 2006

vsage

I don't see you changing your limits of integration when you made the substitution. To address your question though, back up one step and think to yourself "can I make a 2nd substitution that would greatly simplify the problem?".

3. Jun 9, 2006

suspenc3

$$=4\int_{0}^{2\sqrt{3}} 1-cos^2\phi(sin\phi)$$

here?

4. Jun 9, 2006

vsage

Disregarding the missing parentheses, I believe you could do something with that in the form of a substitution (well the same thing is visible in the last one but I felt it was slightly less obvious there)

5. Jun 9, 2006

benorin

A little help

now let $$u=\cos\phi \Rightarrow du=-\sin\phi d\phi$$ and $$0\leq\phi\leq \frac{\pi}{3} \Rightarrow 1\leq u\leq \frac{1}{2}$$ to get

$$=-64\int_{1}^{\frac{1}{2}} (1-u^2)du =64\int_{\frac{1}{2}}^{1} (1-u^2)du$$
$$= 64 \left[ u-\frac{1}{3}u^3\right]_{u = \frac{1}{2}}^{1} = 64 \left[ (1-\frac{1}{3}) - (\frac{1}{2}-\frac{1}{24} )\right]$$
$$=64 \left( \frac{2}{3} - \frac{11}{24}\right) = 64 \frac{5}{24} =\boxed{\frac{40}{3}}$$

The big help here was changing the bounds of integration with each substitution (rather thatn backwards substitution later).

You're doing well: keep going.

-Ben