The value of $\displaystyle \int_0^{\infty}\{\frac{dx}{(a^2+x^2)^7}\}$ is equal to ?

My try:

Put $\displaystyle x=a tan(\theta)$

$\displaystyle dx =asec^2(\theta)d\theta$

Thus integration becomes

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{a sec^2(\theta)d\theta}{a^{14}~(1+tan^2(\theta))^7}$

Using $\displaystyle 1+tan^2(h) = sec^2(h)$

$\displaystyle \frac{1}{a^{13}}\int_{0}^{\frac{\pi}{2}}\{\frac{d\ theta}{(sec^{12}(\theta)}\}$

$\displaystyle \frac{1}{a^{13}}\int_{0}^{\frac{\pi}{2}}\{cos^{12} (\theta)d\theta\}$

Now here I have used the formula that for even "n"

$\displaystyle \int_{0}^{\pi /2} (cos^{n}(xdx)) = \frac{n-1}{n} \cdot \frac{n-3}{n-2} ....\frac{1}{2}\cdot \frac{\pi}{2}$

$\displaystyle \frac{1}{a^{13}}(\frac{11}{12}\cdot \frac{9}{10}...\frac{1}{2}\cdot \frac{\pi}{2})$

My answer thus was $\displaystyle \frac{363}{2048a^{13}}{\color{red}\,\pi}$

Options are

- $\displaystyle \frac{231}{2048a^{13}}$

- $\displaystyle \frac{235}{2048a^{13}}$

--------------------------

I don't have the answer.

Where did I go wrong.