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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!

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  • $\begingroup$ have you tried that $t/2$ substitution maybe...? $\endgroup$ – Calvin Khor Nov 7 '19 at 11:41
  • $\begingroup$ @CalvinKhor I did, but it yielded a rational function which was too complicated for me to integrate. $\endgroup$ – roman Nov 7 '19 at 11:43
  • $\begingroup$ There seems to be a minor discrepancy between your recursive formula and the one I derived in the answer. I double checked mine. $\endgroup$ – Quanto Nov 7 '19 at 14:28
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    $\begingroup$ Yeah, the coefficients a and b are switched $\endgroup$ – Quanto Nov 7 '19 at 14:31
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    $\begingroup$ I was looking at your derivation. There may be an error when you replace back (1) and (2) $\endgroup$ – Quanto Nov 7 '19 at 14:32
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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, $$ J_n={1\over (n-1)(a^2 + b^2)}\left[ {b\sin x-a\cos x \over (a\sin x + b\cos x)^{n-1}} + (n-2)J_{n-2}\right] $$

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  • $\begingroup$ I like this approach much better than mine, it took me a while to apply integration-by-parts since I'm not very used to moving expressions under the sign of differential. Thank you! $\endgroup$ – roman Nov 7 '19 at 14:51

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