Find the area between the following curves and their asymptotes. The curve is given by the polar equation  $ r = a( \sec \theta + \cos \theta) $
At $\theta = \pi /2, r$ goes to infinity, so asymptotes should be at $\theta = \pi /2$
To find the area I integrated $2$ $ \frac{a^{2}(\sec \theta + \cos \theta)^{2}}{2} $ ($= r^{2}/2  \times 2$, since curve is symmetric about $x$ axis) with respect to $\theta$ from $0$ to infinity and got infinity.  
But answer is given $5 \pi a^{2}/2$
Where did I go wrong$?$

We are given the curve
$$ r = a( \sec \theta + \cos \theta) \tag{1}$$
but we are not given the asymptote. 
It is clearly a vertical asymptote of the form $x=$constant, but what is the constant?
So what is happening to $x=r\cos\theta$ as $\theta\to\frac{\pi}{2}$?
We can find this out by multiplying equation (1) by $\cos\theta$ to obtain
$$ r\cos\theta = a( 1 + \cos^2 \theta) $$
Since $\cos(\pi/2)=0$ we see that as $\theta\to\frac{\pi}{2}$, $r\cos\theta\to a$.
So the vertical asymptote of the curve is $x=a$.
In polar coordinates this is $r\cos\theta=a$
So $r=a\sec\theta$ is the polar equation of the asymptote of the function.
So the area between the function and its asymptote can be found by the following:
$$ \int_0^{\pi/2} a^2(\sec\theta+\cos\theta)^2-a^2\sec^2\theta\,d\theta $$
We shall see that $\dfrac{5\pi a^2}{2}$ is not the correct answer.

\begin{eqnarray}
  \int_0^{\pi/2} a^2(\sec\theta+\cos\theta)^2-a^2\sec^2\theta\,d\theta&=&
  a^2\int_0^{\pi/2}2+\cos^2\theta\,d\theta\\
  &=&a^2\int_0^{\pi/2}2+\frac{1+\cos2\theta}{2}\,d\theta\\
  &=&a^2\left[\frac{5}{2}\theta+\frac{1}{4}\sin(2\theta)\right]_0^{\pi/2}\\
  &=&\frac{5\pi a^2}{4}
  \end{eqnarray}

