
#1
Sep1806, 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] 



#2
Sep1806, 08:40 AM

Mentor
P: 6,044

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]? 



#3
Sep2006, 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] 



#4
Sep2106, 05:38 AM

HW Helper
P: 1,422

arctan integralYou 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 antiderivative of this expression, you can use the substitution: [tex]v = \frac{2}{\sqrt{3}} u[/tex]. Can you go from here? :) 



#5
Sep2106, 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] 


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