How to solve the integral using integration by parts?

In summary, Homework Statement integration by parts can be difficult, but with practice and reference to trig identities it can be done. However, the second part can be tedious.
  • #1
t_n_p
595
0
Homework Statement

Hi, I'm trying to solve
http://img204.imageshack.us/img204/7199/untitledke0.png
in terms of I(n-2) but I'm not exactly sure where to start/what to do :rolleyes:
 
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  • #2
You titled this "integration" by parts. Doesn't that give you a hint where to start?

Since it is much easier to differentiate a power of a function than to integrate, the choice of u= sinn(x) and dv= sin(x) seems simplest.

Since the problem says "in terms of I(n-2)" I suspect you will need to do it twice.
 
  • #3
Dido what Halls said, with tiny corrections:

[tex]u= \sin^{n-1} x, dv = \sin x dx[/tex] is what he might have meant.
 
  • #4
Bad mentor! Giving wrong hints, tut tut...

Hints: only perform by parts once; getting [tex]I(n)[/tex] again on the right hand side is no bad thing; remember that for trig problems you'll usually need to use trig identities at some point.
 
  • #5
I was thinking right- my fingers hit the wrong keys!

(and it is "ditto", not "dido"!)
 
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  • #6
Gib Z said:
Dido what Halls said, with tiny corrections:

[tex]u= \sin^{n-1} x, dv = \sin x dx[/tex] is what he might have meant.

How would I diff u?
 
  • #7
Use the chain rule?
 
  • #8
Let u = (n-1) hence y=[sin(x)]^(u)
then du/dn = 1

but again, unsure how to get dy/du now..
 
  • #9
Instead, try letting u = sin x when differentiating :)
 
  • #10
[tex]\frac{d}{dx}\left(\sin^{n-1}(x)\right) = \left((n-2)\sin^{n-2}(x)\right) \left(\cos(x)\right)[/tex]

Btw, you can't really do integration without being able to do differentiation like the back of your hand -- more practise at that would be helpful.
 
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  • #11
Gib Z said:
Instead, try letting u = sin x when differentiating :)

I made a mistake before when doing chain rule. To prevent confusion, I will initially let f(x)=[sin(x)]^(n-1) and g'(x)=sinx

Then I use chain rule,
now I let u = sinx, hence f(x) = u^(n-1)

dy/du=(n-1)(u^(n-2))
and du/dx = cos(x)

hence dy/dx= (n-1)(sinx^(n-2))(cosx)?
 
  • #12
genneth said:
[tex]\frac{d}{dx}\left(\sin^{n-1}(x)\right) = \left((n-2)\sin^{n-2}(x)\right) \left(-\cos(x)\right)[/tex]

Btw, you can't really do integration without being able to do differentiation like the back of your hand -- more practise at that would be helpful.

Can you please explain how you got that/what steps did I do wrong above?
 
  • #13
t_n_p said:
Can you please explain how you got that/what steps did I do wrong above?

You got it right -- I got it wrong -- looks like I need more practice... :redface:
 
  • #14
t_n_p said:
I made a mistake before when doing chain rule. To prevent confusion, I will initially let f(x)=[sin(x)]^(n-1) and g'(x)=sinx

Then I use chain rule,
now I let u = sinx, hence f(x) = u^(n-1)

dy/du=(n-1)(u^(n-2))
and du/dx = cos(x)

hence dy/dx= (n-1)(sinx^(n-2))(cosx)?

*in imitation of Mr. Burns on the Simpsons* Excellent..
 
  • #15
coolio!
I sub all the relevant values into the "magical integration by parts formula", but the second part seems awfully tedious..

http://img468.imageshack.us/img468/1193/untitledyj7.png

Can (n-1) be taken out as a constant, leaving just the sin and cos terms for me to integrate by parts again?
 
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  • #16
Yes it can, but also it wants you to express cos^2 x as 1-sin^2 x, get the original integral "I" on both sides :)
 
  • #17
ah ok!
After converting (cosx)^2 using the trig identity I get..
http://img521.imageshack.us/img521/7374/untitledfv0.png
Where to from there?
 
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  • #18
t_n_p said:
ah ok!
After converting (cosx)^2 using the trig identity I get..
http://img521.imageshack.us/img521/7374/untitledfv0.png
Where to from there?

careful. you can't separate it like that:

[tex]\int{f(x)*g(x)dx}\ne\int{f(x)dx}*\int{g(x)dx}[/tex]

keep the [tex]sin^{n-2}x[/tex] inside the integral, and multiply it out by [tex]-1+sin^2x[/tex]
 
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  • #19
ah ok, you mean expand to give..
http://img512.imageshack.us/img512/6858/untitleddl8.png

but then what?
 
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  • #20
t_n_p said:
ah ok, you mean expand to give..
http://img512.imageshack.us/img512/6858/untitleddl8.png

but then what?

Remember... your right side right now just has [tex]\int{vdu}[/tex]... but it's supposed to be [tex]uv - \int{vdu}[/tex]

To continue... remember that sums within integrals can be separated out into separate integrals (unlike products)... once that's done, you can substitute your I(n-2)... and use algebra to solve for [tex]\int{sin^n(x)dx}[/tex]
 
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  • #21
t_n_p said:
Can (n-1) be taken out as a constant, leaving just the sin and cos terms for me to integrate by parts again?
Yes it can.
 
  • #22
learningphysics said:
Remember... your right side right now just has [tex]\int{vdu}[/tex]... but it's supposed to be [tex]uv - \int{vdu}[/tex]

To continue... remember that sums within integrals can be separated out into separate integrals (unlike products)... once that's done, you can substitute your I(n-2)... and use algebra to solve for [tex]\int{sin^n(x)dx}[/tex]

Thanks for reminder, almost got carried away

After splitting and converting one of the integrals into terms of (I(n-2)), I get..
http://img205.imageshack.us/img205/7929/untitledyw3.png
But now the integral is on both sides...:confused:

edit: just changed the equation as I realized (n-1) must be multiplied by both the integrals after splitting (hence I added square brackets)..
 
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  • #23
t_n_p said:
Thanks for reminder, almost got carried away

After splitting and converting one of the integrals into terms of (I(n-2)), I get..
http://img205.imageshack.us/img205/7929/untitledyw3.png
But now the integral is on both sides...:confused:

edit: just changed the equation as I realized (n-1) must be multiplied by both the integrals after splitting (hence I added square brackets)..

Treat [tex]\int{sin^n(x)}dx[/tex] as an ordinary variable... you can replace it with a variable if you want (on both sides)... then solve for that variable...
 
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  • #24
Ok, after I do that, I'm confused about the expansion of the second part...

(n-1)[I(n-2)+a], where a is the integral [sin(x)]^n

with the expansion of the first term, i.e. n*[I(n-2)], how do I treat it?
 
  • #25
t_n_p said:
Ok, after I do that, I'm confused about the expansion of the second part...

(n-1)[I(n-2)+a], where a is the integral [sin(x)]^n

with the expansion of the first term, i.e. n*[I(n-2)], how do I treat it?

just treat I(n-2) as a variable... don't do anything with it.
 
  • #26
After expanding and taking all a's to one side and dividing by n I get...

http://img470.imageshack.us/img470/4391/untitledeq4.png
 
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  • #27
t_n_p said:
After expanding and taking all a's to one side and dividing by n I get...

http://img470.imageshack.us/img470/4391/untitledeq4.png

shouldn't the first term in your numerator be: [tex]-cos(x)sin^{n-1}(x)[/tex]. Other than that it looks good to me.
 
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  • #28
my bad, crappy typo.

Does that look decent for a final answer? Not much can be canceled but the "n"...
 
  • #29
t_n_p said:
my bad, crappy typo.

Does that look decent for a final answer? Not much can be canceled but the "n"...

Looks fine as a final answer to me. I'd probably just write (n-1)I(n-2) instead of nI(n-2) - I(n-2).
 
  • #30
Sweet, how would I use that solution to then solve..
http://img507.imageshack.us/img507/6310/untitledgl0.png

The 1/ is throwing me off, I'm thinking let denominator = u, but then I don't use the answer found above :confused:
 
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  • #31
t_n_p said:
Sweet, how would I use that solution to then solve..
http://img507.imageshack.us/img507/6310/untitledgl0.png

The 1/ is throwing me off, I'm thinking let denominator = u, but then I don't use the answer found above :confused:

If you take the denominator to the top, this is the integral for n = -4 (in the form of your previous question).

Your previous solution gives I(n) in terms of I(n-2) (the 'a' is I(n))...

Do you have any ideas how to use your previous solution? Hint pluggin in n=-4 won't help... but n =-2 will.
 
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  • #32
Why -2 though?
 
  • #33
t_n_p said:
Why -2 though?

You want an equation with I(-4) in it... pluggin in n=-2 will give you that because I(n-2) = I(-4),

so that will be an equation with I(-2) and I(-4)... I(-2) is a very simple integral... so you can just solve for I(-4).
 
  • #34
Then I get...

-(1/8)-3I(-4), how do I show that = 4/3?
 
  • #35
t_n_p said:
Then I get...

-(1/8)-3I(-4), how do I show that = 4/3?

I'm getting the entire right side as:

[tex]\frac{1}{\sqrt{8}} - \frac{3}{8}I(-4)[/tex]

not sure if I've made a mistake or not..

calculate the left side... I(-2)
 
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