I have difficulty solving the following curvilinear integral
$$\int_{\gamma} x ds$$ where $\rho = 1+ \cos(\theta)$ with $\theta \in [-\pi,\pi]$
I parametrized the curve with $$ \begin{cases} x= \rho(\theta)\cos(\theta) = (1+\cos(\theta))\cos(\theta) \\ y= p(\theta)\sin(\theta) = (1+\cos(\theta))\sin(\theta) \end{cases}$$
As a first step, i calculated the "ds" in this way
$ds = \sqrt{[\rho(\theta)]^2 + [\rho'(\theta)]^2)} d\theta$ where $\rho' = -sin(\theta)$
So $ds = \sqrt{[1+\cos(\theta)]^2 + [-\sin(\theta)]^2)} d\theta = \sqrt{2(1+\cos(\theta))} d\theta =\sqrt{2} \sqrt{1+\cos(\theta)} d\theta$
Then I calculated : $$\int_{\gamma} x ds = \int_{-\pi}^{\pi} [\cos(\theta)(1+\cos(\theta)] \sqrt{2} \sqrt{1+\cos(\theta)} d\theta$$
I don't know if I've done the steps well up to now, but I'm stuck at this integral that I can't approach
Can you give me support please?
Thank you for the hint. $$$$ placing $$1+\cos(\theta) = 2\cos^2(\frac{\theta}{2}) : $$
$$\int_{-\pi}^{\pi} [\cos(\theta)(1+\cos(\theta)] \sqrt{2} \sqrt{1+\cos(\theta)} d\theta = \sqrt{2} \int_{-\pi}^{\pi} (2\cos^2(\frac{\theta}{2})-1)(2\cos^2(\frac{\theta}{2})\sqrt{2\cos^2(\frac{\theta}{2})}d\theta = 2\int_{-\pi}^{\pi}4\cos^5(\frac{\theta}{2})-2\cos^3(\frac{\theta}{2})d\theta$$ Now is simple integral.
It's correct?
$\textbf{Hint}$:
$$1+\cos\theta = 2\cos^2\frac{\theta}{2}$$
Can you proceed from here?