integ by parts Q (1 Viewer)

[mojako]

Let I_n = integral of sin^n (x) dx from x=0 to x=pi/2, where n is an integer, n >= 0.
Using integration by parts, show that for n >= 2,
I_n = (n-1)/n * I_(n-2)

Thanks.

 

[McLake]

Let sinx = s, cosx = c

I_n = I s^n dx
= I s^(n-2)*s^2 dx
= I s^(n-2)(1 - c^2) dx
= I s^(n-2) - s^(n-2)c^2 dx
= I_(n-2) - s^(n-2)c^2 dx

So u = c
du = - s
v = 1/n-1*s^(n-1)
dv = s^(n-2)*c

so I_n = I_(n-2) - [1/(n-1)*s^(n-1)*c + I 1/(n-1)*s^n]
I_n*(n - 1) = I_(n - 2)*(n - 1) - [s^(n-1)*c] - I_n
n*I_n = I_(n - 2)*(n - 1) - (s^(n-1)*c)
I_n = (n - 1)/n*I_(n - 2) - 1/n[s^(n - 1)*c] (n != 0)

sub in the 0 and pi/2 (should have done it earlier, but it dosn't matter)

I_n = (n - 1)/n*I_(n - 2) - 1/n[0 - 0]
I_n = (n - 1)/n*I_(n - 2)

EDIT: Note, the same trick can be used for (cosx)^n

 

[mojako]

oh, thanks McLake ^_^

 

[Jase]

Or at the beginning you can split the (Sinx)^n into Sin^n-1(x) . Sin(x)
and let dv/dx = sin(x) and carry on from there.
Totally useless since its about exactly the same as McLake's, but i just thought i'd let you know.

 

[mojako]

Can you give a hint for part (b) (iii)?
Thank you.

 

[mojako]

Or at the beginning you can split the (Sinx)^n into Sin^n-1(x) . Sin(x)
and let dv/dx = sin(x) and carry on from there.
Totally useless since its about exactly the same as McLake's, but i just thought i'd let you know.

I tried it
but it gave me I_n (as in.. I_n = I_n, which is true but useless...)

 

[Jase]

really?

well okay, you have :

u = (sinx^)n-1 u`= (n-1)cosx (sinx)^n-2
v = -cosx v`= sinx

I-n = [-cosx(sinx)^n-1]pi/2 -> 0 + (n-1)Int{ (cosx)^2(sinx)^n-2 dx

replacing cosx^2 with 1 - sinx^2 :
I-n = (n-1) Int { (1 - sin^2(x))(sinx)^n-2 dx
= (n-1) Int{ sin^n-2(x) - sin^n(x) dx
hence I-n = (n-1).I_(n-2) - (n-1).I_n
then (1 + n -1)I_n = (n-1). I_(n-2)
I_n = (n-1)/n . I_n-2) qed

 

[Jase]

Can you give a hint for part (b) (iii)?
Thank you.

The only thing i have to say for that is.. its hard.

first, start with the result in part a(iv), and root everything.. and flip.
then try and make that middle part equal to P_n and keep moving stuff around.

 

[mojako]

ooh...
I did "(n-1)Int{ (cosx)^2(sinx)^n-2 dx" using integ by parts like in McLake solution (so I did integrations by parts 2 times, which return me to I_n)

Part (iii) isn't tough with your hint ^^