Eh guys, recently i was practicing reduction formulae questions and i stumbled on a problem which adds more problem to me. :) I was unable to do it.

Find the reduction formulae of

\(I_{n}\) = \(\displaystyle\int_{0}^{\pi} e^{-x} \sin^{n}x dx\)

pls guys, help me!!!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHere's the solution

Using Integration By Parts, we get

\[I_n = \int_0^\pi \sin^n x\ e^{-x}\ dx \] \[= \left(\sin^n x \int e^{-x}\ dx\right|_0^\pi \ + \ n\int_0^\pi \sin^{n-1}x \cos x\ e^{-x}\ dx\] \[= n\int_0^\pi \sin^{n-1}x \cos x\ e^{-x}\ dx\]

Again applying Integration By Parts

\[= n\left(\sin^{n-1} x \cos x \int e^{-x}\ dx\right|_0^\pi \ +\ n\int_0^\pi \left((n-1)\sin^{n-2} x \cos^2 x - \sin^n x\right)e^{-x}\ dx\] \[= n\int_0^\pi \left((n-1)\sin^{n-2} x (1-\sin^2 x) - \sin^n x\right)e^{-x}\ dx\] \[= n\int_0^\pi \left((n-1)\sin^{n-2} x - n\sin^n x\right)e^{-x}\ dx\] \[= n(n-1)\int_0^\pi \sin^{n-2} x\ e^{-x}\ dx - n^2 \int_0^\pi \sin^n x\ e^{-x}\ dx\] \[= n(n-1) I_{n-2} - n^2I_n = I_n\]

On simplifying, we get

\[\boxed{I_n = \dfrac{n(n-1)}{n^2+1}I_{n-2}}\]

Now, we compute, \(I_0\) and \(I_1\) using successive IBP.

\[I_0 = \int_0^\pi e^{-x}\ dx = 1-e^{-\pi}\]

\[\text{and}\ I_1 = \int_0^\pi e^{-x} \sin x\ dx = \int_0^\pi e^{-x} \cos x\ dx \\ = \left(-\cos x e^{-x} \right|_0^\pi + \int_0^\pi \sin x\ e^{-x}\ dx = (1+e^{-\pi}) + I_1 \\ \Rightarrow I_1 = \dfrac{1+e^{-\pi}}{2}\]

So, can you guess, why did we compute \(I_0\) and \(I_1\)?

It's because if \(n\) is even then -

\[I_n = \dfrac{n(n-1)}{n^2+1}I_{n-2} = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}I_{n-4} =\cdots = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}\ldots I_0\]

and similarly if \(n\) is odd, then

\[I_n = \dfrac{n(n-1)}{n^2+1}\dfrac{(n-2)(n-3)}{(n-2)^2+1}\ldots I_1\]

As an example, we try computing the integral for \(n=2,3,\ldots\)

\[I_2 = \frac{2}{5}I_0 = \frac{2}{5}(1-e^{-\pi})\]

\[I_3 = \frac{3}{5}I_1 = \frac{3}{10}(1+e^{-\pi})\]

\[I_4 = \frac{8}{17}I_2 = \frac{16}{85}(1-e^{-\pi})\]

And now, similarly, you can compute the integral for any \(n \in N\)

Note: I am not sure if an explicit solution to the above recurrence, that we found, exists or not and I am still working on it's solution. But W|A gives the solution to that recurrence formula as

\[I_n = \dfrac{2\pi c_1 \text{csch}(\pi) \Gamma(n) \Gamma(n+1)}{\Gamma(n+1-i) \Gamma(n+1+i)}\]

Log in to reply

thanks so much. initially i used integration by parts but i got confused at a point. Thanks once again.

Log in to reply

It was all mine pleasure.

Log in to reply