MHB Solving a Difficult Integral: Evaluating \int_{C}^{} \,d \frac{x^3}{y} ds

[AngryStyro]

Hi, I just wonder if I'm doing this question correctly, as it seems like I'm stuck with a particularly difficult integral.

Let C be the curve $$ y = \frac{x^2}{2} , 0 \le x \le 2 $$ , evaluate $$ \int_{C}^{} \,d \frac{x^3}{y} ds $$

So I have so far:

Parameterise C as $$ r(t) = (x(t),y(t)) = (t, \frac{t^2}{2} ) , 0 \le t \le 2$$

Then we have:

$$ x = t, \d{x}{t} = 1 $$
$$ y = \frac{t^2}{2} , \d{y}{t} = t $$

$$ ds = \sqrt{(1+t^2)} dt $$

So $$ \int_{C}^{} \, \frac{x^3}{y} ds = \int_{0}^{2} \, \frac{(t^2)^3}{0.5t^2} \sqrt{1+t^2} dt $$
$$ = \int_{0}^{2} \, 2t^4 \sqrt{1+t^2} dt $$

..there doesn't really seem like an easy way to solve this integral, which is why I think I have gone wrong somewhere?

 

[Opalg]

$$ x = t, \d{x}{t} = 1 $$
$$ y = \frac{t^2}{2} , \d{y}{t} = t $$

$$ ds = \sqrt{(1+t^2)} dt $$

So $$ \int_{C}^{} \, \frac{x^3}{y} ds = \int_{0}^{2} \, \frac{\color{red}{(t^2})^3}{0.5t^2} \sqrt{1+t^2} dt $$
$$ = \int_{0}^{2} \, 2t^4 \sqrt{1+t^2} dt $$
..there doesn't really seem like an easy way to solve this integral, which is why I think I have gone wrong somewhere?

Shouldn't that $\color{red}{(t^2})^3$ be $t^3$?

 

[AngryStyro]

Yes, whoops. So that gives:

So $$ \int_{C}^{} \, \frac{x^3}{y} ds = \int_{0}^{2} \, \frac{t^3}{0.5t^2} \sqrt{1+t^2} dt $$
$$ = \int_{0}^{2} \, 2t \sqrt{1+t^2} dt $$

It is still a rather difficult integral to evaluate though...

Never mind, I see that I can solve with a simple substitution now (Angry)

Thankyou!