1. ## Trig Integral Questions

Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.

First one:

∫ dx/[x*√(5-x^2)]

Second one:

∫ x^3*√(9-x^2) dx

Thanks

2. Originally Posted by SportfreundeKeaneKent
Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.

Second one:

∫ x^3*√(9-x^2) dx

Thanks
Make the substitution $\displaystyle u = \sqrt{9 - x^2} \Rightarrow x^2 = 9 - u^2$ and $\displaystyle dx = - \frac{\sqrt{9 - x^2}}{x} \, du$.

3. Originally Posted by SportfreundeKeaneKent
Can't solve these two. Apparently you're NOT supposed to use a trig substitution for the second one. I don't know how you can for the first since 5 isn't a prime number.

First one:

∫ dx/[x*√(5-x^2)]

[snip]
Do you have to use a trig substitution?

Life's much easier when you make the substitution $\displaystyle u = \sqrt{5 - x^2} \Rightarrow x^2 = 5 - u^2$ and $\displaystyle dx = - \frac{\sqrt{5 - x^2}}{x} \, ....$

4. I don't see how the second one works with the substitution.

I can get to x/√5-x^2 dx but then I'm stuck.

I pretty much don't understand why x's are still in there if I'm trying to put all u's into the equation.

5. For ∫ x^3*√(9-x^2) dx

Mr. Fantastic told you what to substitute. If you take $\displaystyle u = \sqrt{9-x^2}$.

$\displaystyle du = \frac{-\not{2}x}{\not{2}\sqrt{9-x^2}} dx$

$\displaystyle du = -\frac{x}{\sqrt{9-x^2}} dx$

$\displaystyle dx = -\frac{\sqrt{9-x^2}}{x} du$

Now replace! you'll see that you can substitute $\displaystyle 9-x^2$ and $\displaystyle x^2$ (remember Mr. F said $\displaystyle u^2 = 9 - x^2$).