Integral (3)

Problem:

$\int\frac{dt}{P + Qt^2}$


Solution:

Let $\sqrt{\frac{P}{Q}}y = t$, so $\sqrt{\frac{P}{Q}}dy = dt$.


$\begin{eqnarray*} & & \int\frac{dt}{P + Qt^2} \\ &=& \int\frac{\sqrt{\frac{P}{Q}}dy}{P + Q\left(\sqrt{\frac{P}{Q}}y\right)^2} \\ &=& \int\frac{\sqrt{\frac{P}{Q}}dy}{P + Q\frac{P}{Q}y^2} \\ &=& \int\frac{\sqrt{\frac{P}{Q}}dy}{P + Py^2} \\ &=& \sqrt{\frac{P}{Q}}\int\frac{dy}{P + Py^2} \\ &=& \sqrt{\frac{1}{PQ}}\int\frac{dy}{1 + y^2} \\ &=& \sqrt{\frac{1}{PQ}}\tan^{-1} y \\ &=& \sqrt{\frac{1}{PQ}}\tan^{-1} \left(\sqrt{\frac{Q}{P}}t\right) \\ \end{eqnarray*}$