Arc Length of y^2=4(x+4)^3 from x=0 to x=2

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Homework Statement


Find the arc length of the equation y^2=4(x+4)^3 from x=0 to x=2

Homework Equations


L=\int_{a}^{b}\sqrt{1+f'(x)}dx


The Attempt at a Solution


L=\int_{0}^{2}\sqrt{1+9(x+4)}dx
which simplifies in to
L=\int_{0}^{2}\sqrt{9x+37}dx
and I'm stuck there--how should i try to integrate that?
 
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Substitute u=9x+37?
 
oh. wow. thanks.
now i feel kinda dumb lol i was making it more complicated than i had to, trying trig sub and stuff.
so dx=\frac{du}{9}.
sweet.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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