Okay, I am completely stumped on this problem. Do I have to use tan^-1 somehow?
dy
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y(√5y^2-3)
Thanks guys! this is the last one
Have you learned partial fractions?
$\displaystyle \int \frac{dy}{y(\sqrt 5y^2 - 3)} = \int \frac Ay + \frac {By + C}{\sqrt 5y^2 - 3}$
$\displaystyle A(\sqrt 5y^2 - 3) + By^2 + Cy = 1$
Using y = 0, we isolate A
$\displaystyle A = -\frac 13$
$\displaystyle -\frac 13(\sqrt 5y^2 - 3) + By^2 + Cy = 1$
Play some algebra tricks
$\displaystyle 3By^2 + 3Cy - (\sqrt 5y^2 - 3) = 3$
$\displaystyle (3B - \sqrt 5)y^2 + 3Cy = 6$
$\displaystyle (3B - \sqrt 5) + \frac{3C}y = \frac 6{y^2}$
Use y = 0 again
$\displaystyle B = \frac{\sqrt 5}3$
From here we've completed the decomposition, C will just be 0.
$\displaystyle \frac 13\int \frac {\sqrt 5y}{\sqrt 5y^2 - 3} - \frac 1{y}$
Use a u-sub
$\displaystyle \frac 16 ln(\sqrt 5y^2 - 3) - \frac 13 ln(y)$
Did I copy the original question correctly? You can work backwards and see that your teacher's answer is clearly not the same integral.
$\displaystyle x = \frac {\sqrt 3}3tan^{-1}(\frac{\sqrt 3}3(5y^2-3))$
$\displaystyle tan(\frac 3{\sqrt 3}x) = \frac{\sqrt 3}3(5y^2-3)$
$\displaystyle tan(\frac 3{\sqrt 3}x) = \frac{5\sqrt 3}3y^2-\sqrt 3$
$\displaystyle \frac 3{\sqrt 3}sec^2(\frac 3{\sqrt 3}x)\frac{dx}{dy} = \frac{10\sqrt 3}3y$
$\displaystyle \frac 3{\sqrt 3}(tan^2(\frac 3{\sqrt 3}x) + 1)\frac{dx}{dy} = \frac{30}9y$
$\displaystyle \frac{dx}{dy} = \frac{\frac{10\sqrt 3}3y}{\frac 3{\sqrt 3}(tan^2(\frac 3{\sqrt 3}x) + 1)}$
$\displaystyle \frac{dx}{dy} = \frac{10\sqrt 3y}{\frac 9{\sqrt 3}(\frac{5\sqrt 3}3y^2-\sqrt 3)^2 + \frac 9{\sqrt 3}}$
$\displaystyle \frac{dx}{dy} = \frac{10\sqrt 3y}{\frac 9{\sqrt 3}(\frac{75}9y^4 - 10y^2 + 3) + \frac 9{\sqrt 3}}$
$\displaystyle \frac{dx}{dy} = \frac{30y}{(75y^4 - 90y^2 + 27) + 1}$