terrylee reduction question (1 Viewer)

[coeyz]

In = integral (x -> 0) (1+t^2)^n dt for n = 1,2,3 ...
show that In = 1/(2n+1) (1+x^2)^n x + 2n/(2n+1) (In-1)

THANKS HEAPSS

 

[Timothy.Siu]

S(1+t^2)^n dt

u=(1+t²)ⁿ
v'=1
u'=2tn(1+t²)^n-1
v=t

In=t(1+t²)ⁿ-2n S t²(1+t²)^n-1 dt
=t(1+t²)ⁿ-2n S (t²(1+t²)^n-1 +(1+t²)^n-1-(1+t²)^n-1) dt

=t(1+t²)ⁿ-2n S (1+t²)ⁿ dt + 2n S (1+t²)^n-1 dt

=t(1+t²)ⁿ- 2n I_n+2n I_n-1

I_n= t(1+t²)ⁿ/(2n+1)+2n/(2n+1) I_n-1

 

[coeyz]

also this one**

integral (pi/2 -> 0 ) sin^nx cos^2x dx

show that In = (n-1)/(n+2) In-2

 

[Timothy.Siu]

also this one**

integral (pi/2 -> 0 ) sin^nx cos^2x dx

show that In = (n-1)/(n+2) In-2

S sin^n x cos^2 x dx=S sin^n x cosx cos x dx
u=cos x v'= sin^n x cos x
u'=-sin x v=sin^(n+1) x/(n+1)

In= [sin^(n+1) x/(n+1) . cos x]pi->0 + 1/(n+1)S sin^n x sin^2 x dx
=0+1/(n+1) S sin^n x (1-cos^2 x) dx
=-1/(n+1) S sin^n . cos^2 x dx+ 1/(n+1) S sin^n x dx

I_n=-1/(n+1) I_n + 1/(n+1) S sin^(n-2) x - 1/(n+1) I_n-2

cant do it lol sorry

 

[azureus88]

[maths]\int_{0}^{\pi/2}\sin^nx\cos^2xdx[/maths]

let u=sinx
du=cosxdx
x=0, u=0
x=pi/2, u=1

[maths]I_n=\int_{0}^{1}u^n\sqrt{1-u^2}du\\=\int_{0}^{1}u^{n-1}(u\sqrt{1-u^2})du\\=\int_{0}^{1}u^{n-1}\frac{\mathrm{d} }{\mathrm{d} u}(\frac{-(1-u^2)^\frac{3}{2})}{3})du\\=0+\frac{n-1}{3}\int_{0}^{1}u^{n-2}(1-u^2)^\frac{3}{2}du\\=\frac{n-1}{3}\int_{0}^{1}u^{n-2}(1-u^2)\sqrt{1-u^2}du\\=\frac{n-1}{3}I_{n-2}-\frac{n-1}{3}I_n[/maths]

Rearranging, [maths]I_n=\frac{n-1}{n+2}I_{n-2}[/maths]

 

[Timothy.Siu]

[maths]\int_{0}^{\pi/2}\sin^nx\cos^2xdx[/maths]

let u=sinx
du=cosxdx
x=0, u=0
x=pi/2, u=1

[maths]\int_{0}^{1}u^n\sqrt{1-u^2}du\\=\int_{0}^{1}u^{n-1}(u\sqrt{1-u^2})du\\=\int_{0}^{1}u^{n-1}\frac{\mathrm{d} }{\mathrm{d} u}(\frac{-(1-u^2)^\frac{3}{2})}{3})du[/maths]

Its too tedious typing the rest, but it should be pretty straight forward now. Its kinda similiar to the first part.

ur working out is wrong and u read the question wrong lol

S sin^n x cos x dx is just sin^(n+1)/(n+1) lol

 

[azureus88]

nah, i didnt read it wrong, that cosx part was a typo. It's been fixed.

 

[Timothy.Siu]

nah, i didnt read it wrong, that cosx part was a typo. It's been fixed.

ah i see, my bad

 

[jet]

Phew! I thought I was going down the wrong road with that one. Turns out i wasnt.

 

[Timothy.Siu]

too good

 

[coeyz]

oo thanks a lot love you guys as always!