arctan integral


by jesuslovesu
Tags: arctan, integral
jesuslovesu
jesuslovesu is offline
#1
Sep18-06, 08:10 AM
P: 202
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
[tex]
\frac{3/2}{u^2+3/4}
[/tex]

2
[tex]3/2 * \frac{1}{(4u^2+3)/4}[/tex]

3
[tex]3/2*4 * \frac{1}{4u^2+3}[/tex]

4
[tex]6* \frac{1}{3(4/3u^2+1)}[/tex]
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George Jones
George Jones is offline
#2
Sep18-06, 08:40 AM
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P: 6,044
Quote Quote by 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
[tex]
\frac{3/2}{u^2+3/4}
[/tex]

2
[tex]3/2 * \frac{1}{(4u^2+3)/4}[/tex]

3
[tex]3/2*4 * \frac{1}{4u^2+3}[/tex]

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

Careful. You know that

[tex]\int{ \frac{1}{v^2 + 1} dv = arctan(v) + c.[/tex]

So, for your question, what is [itex]v[/itex]?
istevenson
istevenson is offline
#3
Sep20-06, 08:42 PM
P: 10
something was wrong in step 4.
becuse
[tex]\int { \frac{1} {px^2+1} dx = \frac{1} {\sqrt{p}} arctan(\sqrt{p} x) +c [/tex]
here p > 0.

thus,the correct answer is
[tex]\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c[/tex]

VietDao29
VietDao29 is offline
#4
Sep21-06, 05:38 AM
HW Helper
VietDao29's Avatar
P: 1,422

arctan integral


Quote Quote by 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
[tex]
\frac{3/2}{u^2+3/4}
[/tex]

2
[tex]3/2 * \frac{1}{(4u^2+3)/4}[/tex]

3
[tex]3/2*4 * \frac{1}{4u^2+3}[/tex]

4
[tex]6* \frac{1}{3(4/3u^2+1)}[/tex]
Fine still step 4. :)
You know that:
[tex]\int \frac{du}{u ^ 2 + 1} = \arctan (u) + C[/tex], right?
And now, you've gotten to:
[tex]\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)}[/tex]
Now, to find the anti-derivative of this expression, you can use the substitution: [tex]v = \frac{2}{\sqrt{3}} u[/tex].
Can you go from here? :)
istevenson
istevenson is offline
#5
Sep21-06, 09:40 AM
P: 10
OK,let`s see it

[tex]\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)}[/tex]
and now we use the substitution: [tex]p = \frac{2}{\sqrt{3}} u[/tex]
then we get this equation: [tex]u =\frac{\sqrt{3}}{2} p[/tex]
so:
[tex]\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[/tex]
finally we get the answer:
[tex]\sqrt{3} arctan(\frac{2u} {\sqrt{3}}) +c[/tex]
VietDao29
VietDao29 is offline
#6
Sep21-06, 05:10 PM
HW Helper
VietDao29's Avatar
P: 1,422
Yup, it's correct, congratulations. :)


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