Deriving Reduction of Order Formula: Krazy G's Yahoo Answers Q

[MarkFL]

Here is the question:

If In = ∫ pi/2 AND 0 x^n sin(x) dx ; show that; In = n(pi/2)^n−1 - n(n-1)In−2 (n =/>2):?


If In = ∫ pi/2 AND 0 x^n sin(x) dx ; show that;

In = n(pi/2)^n−1 - n(n-1)In−2 (n =/>2):

[USE INTEGRATION TABLES!]

It is hard to write this equation out, here are a few notes;
In is not LN it is a capital i with n sitting slightly lower.

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello Krazy G,

We are given:

\(\displaystyle I_n=\int_{0}^{\frac{\pi}{2}} x^n\sin(x)\,dx\) where \(\displaystyle 2\le n\)

and we are asked to derive the given reduction of order formula:

\(\displaystyle I_n=n\left(\frac{\pi}{2} \right)^{n-1}-n(n-1)I_{n-2}\)

To begin, let's use integration by parts where:

\(\displaystyle u=x^n\,\therefore\,du=nx^{n-1}\,dx\)

\(\displaystyle dv=\sin(x)\,dx\,\therefore\,v=-\cos(x)\)

Hence, we may state:

\(\displaystyle I_n=\left.-x^n\cos(x) \right|_{0}^{\frac{\pi}{2}}+n\int_{0}^{\frac{\pi}{2}} x^{n-1}\cos(x)\,dx\)

\(\displaystyle I_n=-\left(\frac{\pi}{2} \right)^n\cos\left(\frac{\pi}{2} \right)+0^n\cos(0)+n\int_{0}^{\frac{\pi}{2}} x^{n-1}\cos(x)\,dx\)

\(\displaystyle I_n=n\int_{0}^{\frac{\pi}{2}} x^{n-1}\cos(x)\,dx\)

Let's use integration by parts again where:

\(\displaystyle u=nx^{n-1}\,\therefore\,du=n(n-1)x^{n-2}\)

\(\displaystyle dv=\cos(x)\,dx\,\therefore\,v=\sin(x)\)

Hence, we may state:

\(\displaystyle I_n=\left.nx^{n-1}\sin(x) \right|_{0}^{\frac{\pi}{2}}-n(n-1)\int_{0}^{\frac{\pi}{2}}x^{n-2}\sin(x)\,dx\)

Observing that \(\displaystyle \int_{0}^{\frac{\pi}{2}}x^{n-2}\sin(x)\,dx=I_{n-2}\) we may write:

\(\displaystyle I_n=n\left(\frac{\pi}{2} \right)^{n-1}\sin\left(\frac{\pi}{2} \right)-n\cdot0^{n-1}\sin(0)-n(n-1)I_{n-2}\)

\(\displaystyle I_n=n\left(\frac{\pi}{2} \right)^{n-1}-n(n-1)I_{n-2}\)

Shown as desired.