1. Reduction Formulae

GRR
Im stuck

$\displaystyle I_n=\int(2-x^2)^n .dx \ \ \ \ \ n\geq0$

(limits are $\displaystyle \sqrt{2}$ and $\displaystyle 0$, i dont know how to put them usin the Latex codes, sorry)

show that

$\displaystyle I_n=\frac{4n}{2n+1}I_{n-1} \ \ \ \ \ n\geq1$

:S

2. Originally Posted by Silver
GRR
Im stuck

$\displaystyle I_n=\int(2-x^2)^n .dx \ \ \ \ \ n\geq0$

(limits are $\displaystyle \sqrt{2}$ and $\displaystyle 0$, i dont know how to put them usin the Latex codes, sorry)

show that

$\displaystyle I_n=\frac{4n}{2n+1}I_{n-1} \ \ \ \ \ n\geq1$

:S
Let $\displaystyle u=2-x^2\Rightarrow{\sqrt{2-u}=x}$

so $\displaystyle dx=\frac{-1}{2\sqrt{2-u}}$

That is how I would start, then move unto parts

What do you mean limits? are you saying this a definite integral?

3. Originally Posted by Mathstud28
What do you mean limits? are you saying this a definite integral?
No I meant the limits we are supposed to sub in at the end to get the first bit of the reduction formulae

(meaning this bit: $\displaystyle \frac{4n}{2n+1}$)

thanks

4. Originally Posted by Mathstud28
Let $\displaystyle u=2-x^2\Rightarrow{\sqrt{2-u}=x}$

so $\displaystyle dx=\frac{1}{\sqrt{2-u}}$

That is how I would start, then move unto parts

What do you mean limits? are you saying this a definite integral?
$\displaystyle u = 2 - x^2$

$\displaystyle x = \sqrt{2 - u}$

$\displaystyle dx = - \frac{1}{2 \sqrt{2 - u}} \ du$

5. Originally Posted by janvdl
$\displaystyle u = 2 - x^2$

$\displaystyle x = \sqrt{2 - u}$

$\displaystyle dx = - \frac{1}{2 \sqrt{2 - u}}$

6. Hi !

Originally Posted by janvdl
$\displaystyle u = 2 - x^2$

$\displaystyle x = \sqrt{2 - u}$

$\displaystyle dx = - \frac{1}{2 \sqrt{2 - u}}$
Originally Posted by Mathstud28
Let $\displaystyle u=2-x^2\Rightarrow{\sqrt{2-u}=x}$

so $\displaystyle dx=\frac{1}{\sqrt{2-u}}$

That is how I would start, then move unto parts

What do you mean limits? are you saying this a definite integral?
You two should add the du part !

$\displaystyle dx=-\frac{\color{red}du}{2\sqrt{2-u}}$

Because when substituting dx in the integral with the formula you both gave, there will be no du !

7. Originally Posted by Moo
Hi !

You two should add the du part !

$\displaystyle dx=-\frac{\color{red}du}{2\sqrt{2-u}}$

Because when substituting dx in the integral with the formula you both gave, there will be no du !
Yeah ..I never ever put the du...it gets me in trouble

8. \displaystyle \begin{aligned} I_{n}&=\int_{0}^{\sqrt{2}}{\left( 2-x^{2} \right)^{n}\,dx} \\ & =\int_{0}^{\sqrt{2}}{(x)'\left( 2-x^{2} \right)^{n}\,dx} \\ & =x\left( 2-x^{2} \right)^{n} \bigg|_{0}^{\sqrt{2}}+2n\int_{0}^{\sqrt{2}}{\Big( 2-\left( 2-x^{2} \right) \Big)\left( 2-x^{2} \right)^{n-1}\,dx} \\ & =4nI_{n-1}-2nI_{n} \\ \implies I_{n}&=\frac{4n}{2n+1}I_{n-1}.\quad\blacksquare \end{aligned}