Circular arc, integration

Question

Using polar coordinates I am trying to integrate circular arc, but my end result is 7.138330888 and not 7.1416.... What am I doing wrong? $$r(\theta) = \sqrt{(2\cdot{sin}(\frac{\frac{\pi}{2}}{a}\cdot\theta)+2)^2+(2\cdot{cos}(\frac{\frac{\pi}{2}}{a}\cdot\theta)+2)^2}$$ $$Area=\frac12\int_0^a r^2 \, d\theta = \frac12\int_0^{0.6435...} (2\cdot{sin}(\frac{\frac{\pi}{2}}{0.6435...}\cdot\theta)+2)^2+(2\cdot{cos}(\frac{\frac{\pi}{2}}{0.6435...}\cdot\theta)+2)^2 \,d\theta = 7.138330888$$

Answer 1

Solving the problem in the efficient way leads to an area equal to $\pi+4$. In the inefficient way, we have to compute $x^2+y^2$ where $(x,y)$ is the largest solution of $y=x \tan\theta$ and $(x-2)^2+(y-2)^2 = 4$, for $\theta\in\left[\arctan\frac{1}{2},\arctan 2\right]$. The area is so given by $$ 2 \int_{\arctan\frac{1}{2}}^{\arctan{2}}\left(\cos\theta+\sin\theta+\sqrt{\sin(2\theta)}\right)^2\,d\theta $$ or, by the substitution $\theta=\arctan u$, $$ 2\int_{1/2}^{2}\left(\frac{1}{1+u^2}+\frac{2\sqrt{2u}+4u+2u\sqrt{2u}}{(1+u^2)^2}\right)\,du.$$ The latter is quite a horrible integral, but it can be computed by partial fraction decomposition as $$ \left[4\arctan(\sqrt{2u}-1)-\frac{4+2\sqrt{2u}(1-u)}{1+u^2}\right]_{1/2}^{2}=\pi+4. $$ The inefficient way leads to an interesting exercise in Calculus.