Integration by parts question

**CourtneyMoon** · October 19th 2011, 07:31 AM

Let Jn= the integral between 0 and 1 of (x^2n)(sin(Pi*x)) .dx where n=1,2,3....

Use integration by parts to show that:
Jn+1 = 1/Pi^2 [Pi - (2n+1)(2n+2) *Jn]

So I started by setting u=sin(Pi*x) and dv=x^(2n+2)
so that du=Pi*cos(Pi*x) and v= (x^(2n+1))/(2n+1)

But this just seems to make the integral harder when I put it together, and so does doing it the other way round. What am I doing wrong? Please could I have some help with this.

**chisigma** · October 19th 2011, 08:08 AM

Originally Posted by CourtneyMoon

Let Jn= the integral between 0 and 1 of (x^2n)(sin(Pi*x)) .dx where n=1,2,3....

Use integration by parts to show that:
Jn+1 = 1/Pi^2 [Pi - (2n+1)(2n+2) *Jn]

So I started by setting u=sin(Pi*x) and dv=x^(2n+2)
so that du=Pi*cos(Pi*x) and v= (x^(2n+1))/(2n+1)

But this just seems to make the integral harder when I put it together, and so does doing it the other way round. What am I doing wrong? Please could I have some help with this.

Is...

$J_{n}= \int_{0}^{1} x^{2n}\ \sin \pi x\ dx = |\frac{x^{2n+1}}{2n+1} \sin \pi x|_{0}^{1} - \frac{\pi}{2n+1}\ \int_{0}^{1} x^{2n+1}\ \cos \pi x\ dx =$

$= -\frac{\pi}{2n+1}\ \{|\frac{x^{2n+2}}{2n+2}\ \cos \pi x|_{0}^{1} + \frac{\pi}{2n+2}\ \int_{0}^{1} x^{2n+2}\ \sin \pi x\ dx \}=$

$= \frac{\pi}{(2n+1)\ (2n+2)}\ (1 - \pi\ J_{n+1})$ (1)

... and reordering (1) You obtain the requested result...

Kind regards

$\chi$ $\sigma$

**Soroban** · October 19th 2011, 09:41 AM

Hello, CourtneyMoon!

I'll get you started . . . I haven't finished it yet.

$\text{Let }\,J_n \;=\; \int^1_0\,x^{2n}\sin(\pi x)\,dx\;\text{ where }n=1,2,3\hdots$

$\text{Use integration by parts to show that:}$
. . $J_{n+1} \;=\;\frac{1}{\pi^2}\bigg[\pi - (2n+1)(2n+2)\!\cdot\!J_n\bigg]$

We have: . $J_{n+1} \;=\;\int^1_0 \! x^{2n+2}\sin(\pi x)\,dx$

. . $\text{Let: }\:\begin{Bmatrix}u &=& x^{2n+2} && dv &=& \sin(\pi x)\,dx \\ du &=& (2n+2)x^{2n+1}dx && v &=& \text{-}\frac{1}{\pi}\cos(\pi x) \end{Bmatrix}$

$J_{n+1} \;=\;\text{-}\frac{1}{\pi}x^{2n+2}\cos(\pi x) + \frac{2n+2}{\pi} \int\! x^{2n+1}\cos(\pi x)\,dx$

. . $\text{Let: }\:\begin{Bmatrix}u &=& x^{2n+1} && dv &=& \cos(\pi x)\,dx \\ du &=& (2n+1)x^{2n}{dx && v &=& \frac{1}{\pi}\sin(\pi x) \end{Bmatrix}$

$J_{n+1} \;=\;\text{-}\frac{1}{\pi}x^{2n+2}\cos(\pi x) \;+\; \frac{2n+2}{\pi}\bigg[\tfrac{1}{\pi}x^{2n+1}\sin(\pi x) \;-\; \frac{2n+1}{\pi}\int\! x^{2n}\sin(\pi x)\,dx\bigg]$

$J_{n+1} \;=\;\text{-}\frac{1}{\pi}x^{2n+2}\cos(\pi x) \;+\; \frac{2n+2}{\pi^2}x^{2n+1}\sin(\pi x) \;-\; \frac{(2n+1)(2n+2)}{\pi^2}\underbrace{\int \! x^{2n}\sin(\pi x)\,dx}_{\text{This is }J_n}$

$J_{n+1} \;=\;\text{-}\frac{1}{\pi}x^{2n+2}\cos(\pi x) + \frac{2n+2}{\pi^2}x^{2n+1}\sin(\pi x) - \frac{(2n+1)(2n+2)}{\pi^2}\cdot J_n\,\bigg]^1_0$

Can you finish up?

**CourtneyMoon** · October 19th 2011, 10:20 AM

Thank you chisigma and Soroban.

Originally Posted by Soroban

$J_{n+1} \;=\;\text{-}\frac{1}{\pi}x^{2n+2}\cos(\pi x) + \frac{2n+2}{\pi^2}x^{2n+1}\sin(\pi x) - \frac{(2n+1)(2n+2)}{\pi^2}\cdot J_n\,\bigg]^1_0$

Can you finish up?

Thanks, I understand what you've done. So I get:
J_n+1 = (1/Pi - (((2n+1)(2n+2))/Pi^2) * J_n) + (((2n+1)(2n+2)/Pi^2) * J_n) but that just leaves me with J_n+1 = 1/Pi :S Can't see what I'm doing wrong.

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