# Arctan integral

1. Sep 18, 2006

### jesuslovesu

Alright so, I'm trying to take the integral of 3/2 / (u^2 + 3/4)

according to the book the answer is sqrt(3) * arctan(2u/sqrt(3))
but when I try to get the integral I get: 2*arctan(2u/sqrt(3)) -- I don't see any way to take a sqrt(3) out of the function.

The graphs look about the same except they are shifted vertically apart. Is this just a constant issue (are the answers basically equivilent)? Or did I mess up my algebra?

1
$$\frac{3/2}{u^2+3/4}$$

2
$$3/2 * \frac{1}{(4u^2+3)/4}$$

3
$$3/2*4 * \frac{1}{4u^2+3}$$

4
$$6* \frac{1}{3(4/3u^2+1)}$$

2. Sep 18, 2006

### George Jones

Staff Emeritus

Careful. You know that

$$\int{ \frac{1}{v^2 + 1} dv = arctan(v) + c.$$

So, for your question, what is $v$?

3. Sep 20, 2006

### istevenson

something was wrong in step 4.
becuse
$$\int { \frac{1} {px^2+1} dx = \frac{1} {\sqrt{p}} arctan(\sqrt{p} x) +c$$
here p > 0.

$$\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c$$

4. Sep 21, 2006

### VietDao29

Fine still step 4. :)
You know that:
$$\int \frac{du}{u ^ 2 + 1} = \arctan (u) + C$$, right?
And now, you've gotten to:
$$\int \frac{6 dx}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 dx}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}$$
Now, to find the anti-derivative of this expression, you can use the substitution: $$v = \frac{2}{\sqrt{3}} u$$.
Can you go from here? :)

5. Sep 21, 2006

### istevenson

OK,let`s see it

$$\int \frac{6 du}{3 \left( \frac{4}{3} u ^ 2 + 1 \right)} = \int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}$$
and now we use the substitution: $$p = \frac{2}{\sqrt{3}} u$$
then we get this equation: $$u =\frac{\sqrt{3}}{2} p$$
so:
$$\int \frac{2 du}{\left( \left( \frac{2}{\sqrt{3}} u \right) ^ 2 + 1 \right)}=\int \frac{2}{p^2+1}d(\frac{\sqrt{3}}{2} p) = \sqrt{3}actan(p) + c$$
$$\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c$$