Evaluating $\oint_C(x-y^3)dx+x^3dy$, where $C$ is the unit circle, in two ways gives two different answers

Question

I'm trying to calculate $\oint_C(x-y^3)dx+x^3dy$ where $C$ is $x^2+y^2=1$ (opposite of clock direction..).
I used $\vec r(t) = (\cos t,\sin t)$ and so $d\vec r=(-\sin t,\cos t)$.
After substituting in the integral I got
$$\int_0^{2\pi}(-\cos t\sin t +\sin^4t+\cos^4t)dt$$
So I decided to deal with $$\begin{align}\sin^4t+\cos^4t&=\frac{1-\sin^2(2t)}{2}+\frac{1+\cos^2(2t)}{2}\\&=\frac{2-\sin^2(2t)+\cos^2(2t)}{2}\\&=\frac{\frac{4-\sin(4t)+\cos(4t)}{2}}{2}\\&=1+\frac{1}{4}\cos(4t)-\frac{1}{4}\sin(4t)\end{align}$$
And if i do the integral I would get $\int_0^{2\pi}1dt=2\pi$. (since all of those $sin,cos$ will give integral of $0$)
But, when I do it this way:
$$\begin{align}\sin^4t+\cos^4t&=(\sin^2t+\cos^2t)^2-2\sin^2t\cos^2t\\&=1-\frac{\sin^2(2t)}{2}\\&=1-\frac{1}{2}\left[\frac{1-\cos(4t)}{2}\right]\\&=\frac{3}{4}+\frac{\cos(4t)}{4}\end{align}$$.
And now I get integral: $\int_0^{2\pi}\frac{3}{4}dt=\frac{3\pi}{2}$

I'm more sure of the second way, but I have checked the first way like 10 times already and I can't see what's the problem, what is making me get different answer than the second way, I can't leave it like this since I might have a real problem right here that I'm not noticing.
I would appreciate any help and feedback! Thanks in advance!

Answer 1

Your second approach is correct. In the first approach, as others said, you have not written $\sin^4 t + \cos^4 t$ correctly.

You can apply Green's theorem alternatively.

$ \vec F = (P, Q) = (x - y^3, x^3)$

$\dfrac{\partial Q}{\partial x} = 3 x^2, \dfrac{\partial P}{\partial y} = - 3y^2$

$\dfrac{\partial Q}{\partial x} - \dfrac{\partial P}{\partial y} = 3 (x^2 + y^2)$

So the line integral is equivalent to

$\displaystyle \int_0^{2\pi} \int_0^1 3 r^3 dr \ d\theta = \dfrac{3\pi}{2}$

Answer 2

Your first formula for $\cos^4t+\sin^4t$ is wrong.

It is not true that: $$\color{red}{\sin^4t=\frac{1-\sin^2{2t}}{2}\\\cos^4t=\frac{1+\cos^22t}2}$$

It is true that:

$$\sin^2t =\frac{1-\cos 2t}2\\\cos^2t=\frac{1+\cos 2t}2$$

So you should get: $$\sin^4t =\left(\frac{1-\cos 2t}2\right)^2\\\cos^4t=\left(\frac{1+\cos 2t}2\right)^2$$ So $$\begin{align}\sin^4t+\cos^{4}{t}&=\frac{1+\cos^22t}{2}\\ &=\frac{1+\frac{1+\cos4t}2}{2}\\&=\frac{3}{4}+\frac{1}{4}\cos 4t \end{align}$$