<s>[INST] Find the area enclosed by the curve $r=2+3\cos \theta$. the question is 
Find the area enclosed by the curve:
$r=2+3\cos \theta$
Here's my steps:
since when $r=0$, $\cos \theta=0$ or $\cos\theta =\arccos(-2/3)$. 
so the area of enclosed by the curve is 2*(the area bounded by $\theta=\arccos(-2/3)$ and $\theta=0$)
the answer on my book is $5\sqrt{5}+（17/2）*\arccos(-2/3)$
I have no idea why there is a $5\sqrt{5}$ since $\arccos(-2/3)=2.300523984$ on my calculator.
 [/INST] Area of curve, $$A=2\int_0^{\arccos(-2/3)}\frac{r^2}{2}d\phi$$ $$\implies A=2\int_0^{\arccos(-2/3)} \frac{(2+3\cos\phi)^2}{2}d\phi$$ $$=2\int_0^{\arccos(-2/3)} \frac{4+9\cos^2\phi+12\cos\phi }{2}d\phi$$ $$=\int_0^{\arccos(-2/3)} (4+9\cos^2\phi+12\cos\phi) d\phi$$ $$=3\sqrt{5}+\frac{17}{2}\cos^{-1}(\frac{-2}{3}).$$
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