How do I find the area of the curve represented by the following equations, $$x=a\cos{t}+\frac{a}{2}\ln{\left(\tan^2{\frac{t}{2}}\right)}\\ y=a\sin t$$
Here's what I tried:
Let $A$ denote the area of the curve then $$A=\int y\,dx\\ and\;dx=\left(-a\sin t + \frac{a}{2}\cot {\frac{t}{2}}.\sec^2 {\frac{t}{2}}\right)dt\\So,\;A=\int (a\sin t)\left(-a\sin t + \frac{a}{2}\cot {\frac{t}{2}}.\sec^2 {\frac{t}{2}}\right)dt $$
Now not only does this seem excessively complicated but also I have no idea within which limits should I integrate this expression. Are there any easier alternatives ?
$\mathbf {P.S}$: Also how would one graph this curve ? What would it look like ?
There are different approaches which lead to the solution. A possibility is to use the symmetric formula $$ A = \tfrac12 \int \mathbf{r} \times d \mathbf{r} =\tfrac12 \int(y dx - x dy)$$ for the evaluation of the area inclosed by the curve. Plugging in the expressions $$ \mathbf{r} = a \bigl[\cos t+\tfrac{1}{2}\ln \tan^2(t/2), \sin t\bigr]^T,$$ $$ d\mathbf{r} = a \cos t \bigl[\cot t, 1 \bigr]^T,$$ you obtain after some simple transformations $$A = - \frac{a^2}4 \int_0^{2\pi} \cos t \,\ln \tan^2(t/2) .$$ Integration by parts (integrating $\cos t$ and taking the derivative of the rest), you obtain $$A= - \underbrace{\frac{a^2}4\sin t \,\ln \tan^2(t/2)\Big|_{t=0}^{2\pi}}_{=0} + \tfrac12 a^2 \int_{0}^{2\pi} dt = \pi a^2.$$