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December 28th, 2014, 11:47 AM   #1
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From: kragujevac

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recursive integration,integration done, how to get formula

Hi there .
I need to integrate function dx/sin^n (x). I did the integration part, i believe it is right.
In=- In-2 -2/(n*sin^n(x)
I am sorry for formula written like this , I am learning coding so I hope I'll soon be able to write it better.
So now i thing the condition should be n>=2 and I started getting formula seperately for odd and even index but I can't see it ?
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December 28th, 2014, 12:40 PM   #2
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$$\newcommand{\d}[1][x]{\,\mathrm d #1}\begin{align*} I_n &= \int \frac{\d}{\sin^n x} = \int \sin^{-n} x \d \\ &= \int \left(1-\cos^2 x\right) \sin^{-n-2} x \d \\ &= I_{n+2} - \int \underbrace{\cos x}_v \, \underbrace{\cos x \sin^{-n-2} x \d}_{\d[u]} & \text{by parts} \\ &= I_{n+2} + \cos x \, \frac{1}{n+1}\sin^{-n-1} x + \int \sin x \, \frac{1}{n+1}\sin^{-n-1} x \d \\ &= I_{n+2} + \frac{1}{n+1}\cos x \, \sin^{-n-1} x + \frac{1}{n+1}I_n \\ \frac{n+2}{n+1}I_n &= I_{n+2} + \frac{1}{n+1}\cos x \, \sin^{-(n+1)} x \\ I_{n+2} &= \frac{n+2}{n+1}I_{n} - \frac{1}{n+1} \cos x \, \sin^{-(n+1)}x \\[12pt] I_{n} &= \frac{n}{n-1}I_{n-2} - \frac{1}{n-1} \cos x \, \sin^{-(n-1)}x \qquad n \gt 1 \end{align*}$$

It's possible that there are sign errors in there.

There is no need for a separate formula for odd and even indices. We simply need to be able to evaluate $I_0$ and $I_1$ by other means (which we can).

$$\begin{align*}I_1 &= \int \frac{\d}{\sin x} = \int \csc x \d \\ &= \int \frac{\csc x ( \csc x + \cot x )}{\csc x + \cot x} \d \\ &= -\int \frac{-\csc^2 x -\csc x \cot x}{\csc x + \cot x} \d \\ &= -\log {\left(A\left| \csc x + \cot x \right|\right)}\end{align*}$$

Last edited by v8archie; December 28th, 2014 at 01:00 PM.
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