Integration problem

**kn336a** · May 19th 2009, 02:45 AM

Okay, I am completely stumped on this problem. Do I have to use tan^-1 somehow?

dy
-----------
y(√5y^2-3)

Thanks guys! this is the last one

**derfleurer** · May 19th 2009, 06:02 AM

Have you learned partial fractions?

$\int \frac{dy}{y(\sqrt 5y^2 - 3)} = \int \frac Ay + \frac {By + C}{\sqrt 5y^2 - 3}$

$A(\sqrt 5y^2 - 3) + By^2 + Cy = 1$

Using y = 0, we isolate A

$A = -\frac 13$

$-\frac 13(\sqrt 5y^2 - 3) + By^2 + Cy = 1$

Play some algebra tricks

$3By^2 + 3Cy - (\sqrt 5y^2 - 3) = 3$

$(3B - \sqrt 5)y^2 + 3Cy = 6$

$(3B - \sqrt 5) + \frac{3C}y = \frac 6{y^2}$

Use y = 0 again

$B = \frac{\sqrt 5}3$

From here we've completed the decomposition, C will just be 0.

$\frac 13\int \frac {\sqrt 5y}{\sqrt 5y^2 - 3} - \frac 1{y}$

Use a u-sub

$\frac 16 ln(\sqrt 5y^2 - 3) - \frac 13 ln(y)$

**kn336a** · May 19th 2009, 01:03 PM

Okay my teacher posted up the answers online and this is what he got.

(√3) tan^-1(√3(5y^2-3)/3)
-----------------------------
3

how did he get to that answer?

**derfleurer** · May 19th 2009, 03:35 PM

Did I copy the original question correctly? You can work backwards and see that your teacher's answer is clearly not the same integral.

$x = \frac {\sqrt 3}3tan^{-1}(\frac{\sqrt 3}3(5y^2-3))$

$tan(\frac 3{\sqrt 3}x) = \frac{\sqrt 3}3(5y^2-3)$

$tan(\frac 3{\sqrt 3}x) = \frac{5\sqrt 3}3y^2-\sqrt 3$

$\frac 3{\sqrt 3}sec^2(\frac 3{\sqrt 3}x)\frac{dx}{dy} = \frac{10\sqrt 3}3y$

$\frac 3{\sqrt 3}(tan^2(\frac 3{\sqrt 3}x) + 1)\frac{dx}{dy} = \frac{30}9y$

$\frac{dx}{dy} = \frac{\frac{10\sqrt 3}3y}{\frac 3{\sqrt 3}(tan^2(\frac 3{\sqrt 3}x) + 1)}$

$\frac{dx}{dy} = \frac{10\sqrt 3y}{\frac 9{\sqrt 3}(\frac{5\sqrt 3}3y^2-\sqrt 3)^2 + \frac 9{\sqrt 3}}$

$\frac{dx}{dy} = \frac{10\sqrt 3y}{\frac 9{\sqrt 3}(\frac{75}9y^4 - 10y^2 + 3) + \frac 9{\sqrt 3}}$

$\frac{dx}{dy} = \frac{30y}{(75y^4 - 90y^2 + 27) + 1}$

**kn336a** · May 19th 2009, 03:56 PM

For the equation I wrote down, the square root is supposed to go across 5y^2-3 entirely as opposed to just 5y^2.

We haven't learned partial fractions yet. I think he just wants u-substitution?

**derfleurer** · May 19th 2009, 04:10 PM

So that's

$\int \frac{dy}{y\sqrt{5y^2-3}}$

If so, than yes, i'm getting an arctan, but still not quite this answer

$\frac{\sqrt 3}3tan^{-1}(\frac{\sqrt 3}3(5y^2-3))$

**kn336a** · May 19th 2009, 04:20 PM

yea, maybe he made a mistake; he does that a lot actually. How did you work through that problem? Thanks for all your help by the way.

**kn336a** · May 19th 2009, 04:24 PM

Actually, your answer is the right answer. I just typed the answer in incorrectly.

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