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

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The discussion focuses on deriving the reduction of order formula for the integral I_n = ∫(0 to π/2) x^n sin(x) dx, specifically showing that I_n = n(π/2)^(n-1) - n(n-1)I_(n-2) for n ≥ 2. The derivation employs integration by parts twice, first to express I_n in terms of an integral involving cos(x) and then to relate it back to I_(n-2). The calculations confirm that the boundary terms vanish, leading to the desired formula. The thread emphasizes the importance of using integration tables and proper notation throughout the derivation. The formula is successfully shown as required.
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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.
 
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Hello Krazy G,

We are given:

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

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

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

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

$$u=x^n\,\therefore\,du=nx^{n-1}\,dx$$

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

Hence, we may state:

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

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

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

Let's use integration by parts again where:

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

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

Hence, we may state:

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

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

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

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

Shown as desired.
 
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