Pedal form to polar form of an ellipse How do I convert the equation
$$\frac{a^2b^2}{p^2}=a^2+b^2-\frac{1}{u^2}$$
into the following equivalent form?
$$u^2=\frac{\sin^2 \theta}{b^2}+\frac{\cos^2 \theta}{a^2}$$
where $$\frac{1}{p^2}=u^2+\left( \frac{du}{d\theta} \right)^2$$
EDIT:
I have tried and found out
$$\frac{uab}{\sqrt{u^2(a^2+b^2)-1-u^4a^2b^2}}\, du = d\theta$$
How do I integrate and express the result in terms of $\sin \theta$ and $\cos \theta$?
$p$ is the perpendicular distance from $O$ to the tangent line to $C$ at the point in case of pedal equation of a curve.  It is converted to $u$ as stated above.
 A: Assume $a>b>0$,
\begin{align}
  t &= \sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta} \\
  t^2 &= (a^2-b^2)\sin^2 \theta+b^2 \\
  \sin^2 \theta &= \frac{t^2-b^2}{a^2-b^2} \\
  t^2 &= a^2+(b^2-a^2)\cos^2 \theta \\
  \cos^2 \theta &= \frac{a^2-t^2}{a^2-b^2} \\
  dt &=
 \frac{(a^2-b^2)\sin \theta \cos \theta}
      {\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \, d\theta \\
  &=\frac{\sqrt{(t^2-b^2)(a^2-t^2)}}{t} \, d\theta \\
  d\theta &= \frac{t \, dt}{\sqrt{(t^2-b^2)(a^2-t^2)}} \\
\end{align}
Now
\begin{align}
  d\theta &= \frac{abu}{\sqrt{(a^2 u^2-1)(1-b^2 u^2)}} \, du \\
  &= \frac{abu \times (ab\, du)}{\sqrt{(a^2 b^2 u^2-b^2)(a^2-a^2b^2u^2)}} \\
  &= \frac{t \, dt}{\sqrt{(t^2-b^2)(a^2-t^2)}} \\
  dt &= ab \, du
\end{align}
With boundary conditions:


*

*$\theta=0$ , $u=\dfrac{1}{r}=\dfrac{1}{b}$

*$\theta=\dfrac{\pi}{2}$ , $u=\dfrac{1}{r}=\dfrac{1}{a}$


Therefore,
$$\fbox{$u=\frac{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}}{ab}$}$$

Alternatively,
  \begin{align}
  \int \frac{t\, dt}{\sqrt{(a^2-t^2)(t^2-b^2)}} &=
  \tan^{-1} \sqrt{\frac{t^2-b^2}{a^2-t^2}} \\
  \int \frac{u\, du}{\sqrt{(a^2 u^2-1)(1-b^2 u^2)}} &=
  \frac{1}{ab} \tan^{-1}
  \left( \frac{b}{a} \sqrt{\frac{a^2 u^2-1}{1-b^2 u^2}} \right)
\end{align}

