Simplifying Complicated Trigonometric Integrals

[ace1412]

Homework Statement

$\int_{C}|y|ds$ where C is the curve $(x^{2}+y^{2})^{2}=2^{2}(x^{2}-y^{2})$

Homework Equations

The Attempt at a Solution

i used polar coordinates $x = r cos \theta$ and $y = r sin \theta$

then substituted into the equation to get $r = 2\sqrt{cos 2\theta}$

since $r\geq0$ gives $-\frac{\pi}{4}\leq \theta\leq \frac{\pi}{4}$

substituting back gives $x = 2\sqrt{cos 2\theta}cos \theta$ and $y = 2\sqrt{cos 2\theta}sin \theta$

then i calculated $\frac{dx}{d\theta}=-\frac{2sin 2\theta cos \theta}{\sqrt{cos 2\theta}}-\frac{2sin \theta}{\sqrt{cos 2\theta}}$ and $\frac{dy}{d\theta}=-\frac{2sin 2\theta sin \theta}{\sqrt{cos 2\theta}}+\frac{2cos \theta}{\sqrt{cos 2\theta}}$

and found $ds=\sqrt{(\frac{dx}{d\theta})^{2}+(\frac{dy}{d\theta})^{2}}d\theta=(4tan 2\theta sec 2\theta+4 cos 2\theta)d\theta$

now substituting back to the integral gives $8\int^{-\frac{\pi}{4}}_{\frac{\pi}{4}}\sqrt{cos 2\theta}|sin \theta|(tan 2\theta sec 2\theta+ cos 2\theta)d\theta$

which looks terribly difficult, so i inferred that i did something wrong somewhere, can someone please shed a bit of light? thanks

 

[jackmell]

Why not just muscle through it? I mean suppose you had to? Could you? Start with this piece:

$$\int \frac{\sin(t) \sin(2t)}{(\cos(2t)^{3/2}}dt$$

How about just that part? Already has the 2t thing in the top and bottom. Maybe start with parts. Keep working through it. See what happens.