Trig substitution

1. Oct 8, 2007

tony873004

[SOLVED] trig substitution

This is from the class notes. Evaluate the integral:
$$\int_{}^{} {\frac{x}{{\sqrt {3 - x^4 } }}dx}$$

$$\begin{array}{l} u = x^2 ,\,\,du = 2x\,dx\,\, \Leftrightarrow \,\,dx = \frac{{du}}{{2x}} \\ \\ \int_{}^{} {\frac{{x^1 }}{{\sqrt {3 - x^4 } }}dx} = \int_{}^{} {\frac{x}{{2x\sqrt {3 - u^2 } }}du} = \int_{}^{} {\frac{1}{{2\sqrt {3 - u^2 } }}du } \\ \end{array}$$

The next step I would want to do using trig substitution is
$$\begin{array}{l} a = \sqrt 3 ,\,x = a\sin \theta = \sqrt 3 \sin \theta \\ \\ \frac{1}{2}\int_{}^{} {\frac{1}{{\sqrt {\sqrt 3 ^2 - \sqrt 3 \sin \theta } }}du} = \frac{1}{2}\int_{}^{} {\frac{1}{{\sqrt {3 - \sqrt 3 \sin \theta } }}du} \\ \end{array}$$

But the next step in the example is:
$$\int_{}^{} {\frac{1}{{2\sqrt {3 - u^2 } }}du = } \frac{1}{2}\sin ^{ - 1} \frac{u}{{\sqrt 3 }} + C$$
How did he get this? Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 8, 2007

cristo

Staff Emeritus
$$\int\frac{du}{2\sqrt{3-u^2}}=\frac{1}{2}\int\frac{du}{\sqrt{3}\sqrt{1-(u/ \sqrt{3})^2}}$$ then let y=u/sqrt(3) hence dy=du/sqrt(3), so we have $$\frac{1}{2}\int\frac{dy\sqrt{3}}{\sqrt{3}\sqrt{1-y^2}}=\frac{1}{2}\int\frac{dy}{\sqrt{1-y^2}}$$ which is a standard integral.

3. Oct 8, 2007

tony873004

Thanks for the reply. But I don't see how you did this:
$$\int\frac{du}{2\sqrt{3-u^2}}=\frac{1}{2}\int\frac{du}{\sqrt{3}\sqrt{1-(u/ \sqrt{3})^2}}$$

I know you're right, because I picked a random value for u and plugged both formulas into my calculator. But I just don't know how you got from one to the other.

4. Oct 8, 2007

cristo

Staff Emeritus
Ok, well let's just consider the denominator: $2\sqrt{3-u^2}$. This can be rewritten as $$2\sqrt{(\sqrt{3})^2-u^2}=2\sqrt{(\sqrt{3})^2[1-(u/\sqrt{3})^2]}$$ by factoring out \sqrt{3}^2. We can then take this outside the square-root sign to give $$2\sqrt{3}\sqrt{1-(u/\sqrt{3})^2}$$ as required.

Last edited: Oct 8, 2007
5. Oct 8, 2007

tony873004

I was about to write back and ask how you did that factoring, but now that I've stared at it for a few minutes, I see what you did.

They expected me to come up with that on my own??

Thanks, Christo :)

6. Oct 8, 2007

cristo

Staff Emeritus
You're welcome!