Alternative proof of reduction formula for $\int {dx \over (a\sin x + b\cos x)^n }$ 
The problem i was working on is to prove:
  $$
\begin{align}
J_n &= \int {dx \over (a\sin x + b\cos x)^n } \\
&={1\over (n-1)(a^2 + b^2)}\left({b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right)
\end{align}
$$
  Where $a^2+b^2 \ne 0$ and $n>1\in\Bbb N$

It is indeed true, but the method I've used requires proving additional reduction formula and now I'm trying to find a simpler method. Below is what I've used to prove the statement.
I have started with a simpler case, namely $J_1$:
$$
J_1 = \int {dx\over a\sin x + b\cos x}
$$
Using a bit of trigonometry one may show that:
$$
J_1 = \int {dx\over \sqrt{a^2+b^2}\left({a\sin x\over \sqrt{a^2+b^2}} + {b\cos x\over \sqrt{a^2+b^2}}\right)}
$$
Where for some $\phi$:
$$
\begin{align*}
\sin \phi &= {a\over \sqrt{a^2+b^2}}\tag 1\\
\cos \phi &= {b\over \sqrt{a^2+b^2}}\tag 2
\end{align*}
$$
So:
$$
\begin{align}
J_1 &= \int {dx\over \sqrt{a^2+b^2}\left(\sin x\sin\phi + \cos x\cos \phi\right)}\\
&=\int {dx\over \sqrt{a^2+b^2}\cos(x-\phi)}
\end{align}
$$
So going back to $J_n$ we obtain:
$$
\begin{align}
J_n &= \int {dx\over (\sqrt{a^2+b^2})^n\cos^n(x-\phi)} \\
&= \int {1\over (\sqrt{a^2+b^2})^n}{dx\over \cos^n(x-\phi)}\\
&\stackrel{t=x-\phi}{=} {1\over (\sqrt{a^2+b^2})^n} \int {dt\over \cos^n(t)}\\
\end{align}
$$
Some time ago I've proven a reduction formula for integrals of ${1\over \cos^n x}$, so I'm just going to use it without a proof:
$$
J_n = {1\over (\sqrt{a^2+b^2})^n}\left({\sin t\over (n-1)\cos^{n-1}t} + (n-2)\int{dt\over \cos^{n-2}t}\right)
$$
The tricky part here is to arrive to the expression in the problem statement:
$$
J_n = {1\over (n-1)(a^2 + b^2)}{1\over (\sqrt{a^2 + b^2})^{n-2}}\left({\sin(x-\phi)\over \cos^{n-1}(x-\phi)} + (n-2)\int {dt\over cos^{n-2}t}\right)\\
= {1\over (n-1)(a^2 + b^2)}\left({\sin(x-\phi)\over (\sqrt{a^2 + b^2})^{n-2}\cos^{n-1}(x-\phi)} + {(n-2)\over (\sqrt{a^2 + b^2})^{n-2}}\int {dt\over cos^{n-2}t}\right)\\
={1\over (n-1)(a^2 + b^2)}\left({\sin(x-\phi)\over (\sqrt{a^2 + b^2})^{n-2}\cos^{n-1}(x-\phi)} + J_{n-2}\right)
$$
Consider the following part of the expression:
$$
\begin{align}
E &= {\sin(x-\phi)\over (\sqrt{a^2 + b^2})^{n-2}\cos^{n-1}(x-\phi)} \\
&= {\sin x\cos \phi - \cos x\sin \phi\over (\sqrt{a^2 + b^2})^{n-2}(\cos x\cos \phi + \sin x\sin \phi)^{n-1}}
\end{align}
$$
Now replace back using $(1)$ and $(2)$:
$$
E = {b\sin x - a\cos x\over (\sqrt{a^2 - b^2})^{n-1}\left({b\cos x\over \sqrt{a^2+b^2}} + {a\sin x\over \sqrt{a^2+b^2}}\right)^{n-1}} \\
= {b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}}
$$
Finally:
$$
J_n = {1\over (n-1)(a^2 + b^2)}\left({b\sin x - a\cos x\over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right)
$$
I'm wondering if there are alternative ways to prove the statement? Also, I would appreciate it if someone verifies my writings. Thank you!

Note that,
$$\left({a\cos x - b\sin x\over a\sin x + b\cos x} \right)'
=-{a^2+b^2\over (a\sin x + b\cos x)^2} 
$$
and apply the integration-by-parts,
$$J_n = \int {1 \over (a\sin x + b\cos x)^n } $$
$$ = -\frac{1}{a^2+b^2}\int {1 \over (a\sin x + b\cos x)^{n-2} } d\left({a\cos x - b\sin x\over a\sin x + b\cos x} \right)
$$
$$ = -\frac{1}{a^2+b^2}\left[{a\cos x - b\sin x\over (a\sin x + b\cos x)^{n-1}}
 +(n-2)\int {(a\cos x - b\sin x)^2\over (a\sin x + b\cos x)^{n} } dx\right]
$$
Recognize,
$$(a\cos x - b\sin x)^2 = (a^2+b^2) - (a\cos x + b\sin x)^2 $$
to express the integral as,
$$J_n = -\frac{1}{a^2+b^2}\left[{a\cos x - b\sin x\over (a\sin x + b\cos x)^{n-1}}
 +(n-2)(a^2+b^2)J_n - (n-2)J_{n-2} \right]
$$
Rearrange to obtain,
$$
J_n={1\over (n-1)(a^2 + b^2)}\left[ {b\sin x-a\cos x \over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right]
$$
