Integration: Reduction Formula (1 Viewer)

Hello_World2 · Jan 30, 2023

Screen Shot 2023-01-30 at 10.36.55 pm.png

I tried using integration by parts but they still don't work and I don't know how to simplify
pls help

Drongoski · Jan 30, 2023

$I_n = \int ^1 _0 x^n \sqrt{1-x} dx = \frac {-2}{3}\int ^1 _0 x^n d(1-x)^\frac {3}{2} \\ \\ = \frac {-2}{3}\left[x^n (1-x)^{\frac {3}{2}} \right ]^1 _0 + \frac {2}{3}\int ^1 _0 (1-x)^\frac {3}{2} n \cdot x^{n-1}dx \\ \\ = 0 + \frac {2n}{3} \int ^1 _0 (1-x) \sqrt {1-x} \cdot x^{n-1} dx \\ \\ = \frac {2n}{3} \int ^1 _0 (x^{n-1} \sqrt {1-x} - x^n \sqrt {1-x}) dx \\ \\ = \frac {2n}{3} \int ^1 _0 x^{n-1}\sqrt {1-x} \cdot dx - \frac {2n}{3} \int ^1 _0 x^n \sqrt {1-x} \cdot dx \\ \\ = \frac {2n}{3} I_{n-1} - \frac {2n}{3} I_n \\ \\ \therefore (1 + \frac {2n}{3})I_n = \frac {2n}{3} I_{n-1} \\ \\ \therefore I_n = \frac {2n}{2n+3} I_{n-1}$

$\therefore I_n = \frac {2n}{2n+3} I_{n-1}= \frac {2n}{2n+3} \cdot \frac {2(n-1)}{2n+1} I_{n-2} = \cdots$

I haven't the energy to continue.

Hello_World2 · Jan 30, 2023

Drongoski said:
$I_n = \int ^1 _0 x^n \sqrt{1-x} dx = \frac {-2}{3}\int ^1 _0 x^n d(1-x)^\frac {3}{2} \\ \\ = \frac {-2}{3}\left[x^n \sqrt{1-x} \right ]^1 _0 + \frac {2}{3}\int ^1 _0 (1-x)^\frac {3}{2} nx^{n-1}$

That's where I got Up to bruh idk what to do next

yanujw · Jan 30, 2023

Hello_World2 said:
That's where I got Up to bruh idk what to do next

$(1-x)^\frac{3}{2}nx^{n-1}=(1-x)\sqrt{1-x}nx^{n-1}=n\sqrt{1-x}x^{n-1}-nx^{n}\sqrt{1-x}$

Now you have terms involving U(n) and U(n-1) and so you can progress through the reduction formula.

Drongoski · Jan 31, 2023

For Q24 can use:

$d\sqrt{1-x^2} = \frac {-x}{\sqrt{1-x^2}}dx \\ \\ I_{2n+1} = \int^-_- \frac {x^{2n+1}}{\sqrt{1-x^2}} \cdot dx = - \int ^-_- x^{2n}\frac {-x}{\sqrt{1-x^2}}\cdot dx = - \int ^-_- x^{2n} d\sqrt{1-x^2}\\ \\ = -\left [ x^{2n}\cdot \sqrt{1-x^2} \right ] ^1 _0 + \int ^-_- \sqrt {1-x^2}\cdot 2nx^{2n-1} dx\\ \\ = 0 + 2n \int ^-_- x^{2n-1} \frac {(\sqrt{1-x^2})^2}{\sqrt {1-x^2}} dx\\ \\ = 2n\int ^-_- \left [ \frac {x^{2n-1}}{\sqrt {1-x^2}} - \frac {x^{2n+1}}{\sqrt {1-x^2}} \right ] dx \\ \\ = 2nI_{2n-1} - 2nI_{2n+1}\\ \\ \therefore (1+2n)I_{2n+1} = 2nI_{2n-1}\\ \\ \therefore I_{2n+1} = \frac {2n}{2n+1}I_{2n-1}$

For Q24, 2nd part, I'll leave it to others; it's too tedious for a hunt-&-peck typist.

Drongoski · Jan 31, 2023

In using Integration by Parts, I prefer using the form:

$\int udv = uv - \int vdu$

But I think whilst this form looks simpler & cleaner, it requires you to be good with the use of differentials, like:

$d(x+1), d(1 + ln x), d\sqrt{1-x^2} $ or $ dx = d(\sqrt x ^2) = 2\sqrt x\cdot d\sqrt x.$

Because I base much of my work on this approach, I've developed this technique into a fine art.

Make a Difference – Donate to Bored of Studies!

Integration: Reduction Formula (1 Viewer)

Hello_World2

Member

Drongoski

Well-Known Member

Hello_World2

Member

yanujw

Well-Known Member

Drongoski

Well-Known Member

Drongoski

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)