How to solve the integral using integration by parts?

t_n_p
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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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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.
 
Dido what Halls said, with tiny corrections:

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

Hints: only perform by parts once; getting I(n) 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.
 
I was thinking right- my fingers hit the wrong keys!

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

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

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

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

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:
\frac{d}{dx}\left(\sin^{n-1}(x)\right) = \left((n-2)\sin^{n-2}(x)\right) \left(-\cos(x)\right)

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:

\int{f(x)*g(x)dx}\ne\int{f(x)dx}*\int{g(x)dx}

keep the sin^{n-2}x inside the integral, and multiply it out by -1+sin^2x
 
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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 \int{vdu}... but it's supposed to be uv - \int{vdu}

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 \int{sin^n(x)dx}
 
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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 \int{vdu}... but it's supposed to be uv - \int{vdu}

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 \int{sin^n(x)dx}

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 \int{sin^n(x)}dx 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: -cos(x)sin^{n-1}(x). 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:

\frac{1}{\sqrt{8}} - \frac{3}{8}I(-4)

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

calculate the left side... I(-2)
 
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  • #36
hmm, with that 1/root(8)

What is (1/root(2))^-3? Isn't it 2root(2)?
2root(2) * 1/root(2) = -2?
which then becomes -1?
 
  • #37
t_n_p said:
hmm, with that 1/root(8)

What is (1/root(2))^-3? Isn't it 2root(2)?
2root(2) * 1/root(2) = -2?
which then becomes -1?

I'm getting that from calculating -cos(x)sin^{n-2}(x) = -cos(x)sin^{-4}(x) from your formula for I(n).

At x = pi/2, this quantity is 0.

At x = pi/4, this quantity is -cos(\pi/4)sin^{-4}(\pi/4) = -\sqrt{2}/2 * (\sqrt{2}/2)^{-4} = -\frac{1}{(\sqrt{2}/2)^3} = -\frac{8}{\sqrt{8}}

taking the value at pi/2 and subtracting the value at pi/4, I get \frac{8}{\sqrt{8}}

Then dividing by n = 8, I get \frac{1}{\sqrt{8}} or \frac{\sqrt{2}}{4}
 
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  • #38
isnt that that first term to the power (n-1) making the second term to the poewr of -3?
 
  • #39
t_n_p said:
isnt that that first term to the power (n-1) making the second term to the poewr of -3?

I'm using the equation in the second line of post #26 in this thread. the first term is to the power of n-2.
 
  • #40
It was a typo, even you pointed that out :P
 
  • #41
t_n_p said:
It was a typo, even you pointed that out :P

lol! I'm sorry dude! I've got a poor memory! And why the heck did I divide by n=8, n = -2... damn I'm going nuts.

Ok now I'm getting -1. -cos(45)sin^-3(45) = -2

So the difference is 0 - (-2) = 2. Then divide by n=-2, I get -1...

lol! So the right side is -1 + (3/2)I(-4) ??
 
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  • #42
From the pi/2 terminal I get -(3/2)I(-4) and from the pi/4 terminal, 1 + (3/2)I(-4)
Subtracting to give -1 - 3I(-4) on the RHS.
 
  • #43
t_n_p said:
From the pi/2 terminal I get -(3/2)I(-4) and from the pi/4 terminal, 1 + (3/2)I(-4)
Subtracting to give -1 - 3I(-4) on the RHS.

shouldn't it be (3/2)I(-4) for pi/2...
 
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  • #44
yeah my bad, hence the lhs integral = -1
Now I solve the lhs integral?
Do I let n=0 and do it all again?
 
  • #45
t_n_p said:
yeah my bad, hence the lhs integral = -1
Now I solve the lhs integral?
Do I let n=0 and do it all again?

One thing... I(-4) is a function of x. So at pi/2, this really evaluates to I(-4)(pi/2)... and at pi/4 it's I(-4)(pi/4).

So your right side is really:

-1 + 3/2*[I(-4)(pi/4) - I(-4)(pi/2)]

the quantity in the square brackets is what you want to solve for... it is the integral evaluated between pi/4 and pi2.

Calculate the left side: it is I(-2)... you don't need the formula to do this. It is a simple integral.
 
  • #46
ok, you say I(-4) is a function of x, shouldn't that mean I get

(3pi*I(-4)/-4) - (pi/4) if I I put (pi/2)*I(-4) and (pi/4)*I(-4) respectively?
 
  • #47
t_n_p said:
ok, you say I(-4) is a function of x, shouldn't that mean I get

(3pi*I(-4)/-4) - (pi/4) if I I put (pi/2)*I(-4) and (pi/4)*I(-4) respectively?

The I(-4)(pi/4) isn't multiplying by pi/4... it's like f(pi/4)... where f is a function ie: f(x).

You can write the function as I(-4)(x) and then you're evaluating it at pi/4 and pi/2 and subtracting...
 
  • #48
Damn this is a difficult question!

Can I write it as [I(-4pi/4) or does it have to be I(-4)(pi/4)?
 
  • #49
t_n_p said:
Damn this is a difficult question!

Can I write it as [I(-4pi/4) or does it have to be I(-4)(pi/4)?

No I(-4pi/4) isn't right. I think maybe... I(-4)(pi/4) is best or (I(-4))(pi/4)

Or you can write the whole thing like this after getting rid of the I's:

\int_{\pi/4}^{\pi/2}sin^{-2}(x)dx = -1 + (3/2)\int_{\pi/4}^{\pi/2}sin^{-4}(x)dx

You're just putting limits on the functions on both sides... the integrals take on the limits of pi/2 and pi/4 and for the other functions you just evaluate the difference...
 
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  • #50
ok I do that, but my RHS is different to what you suggested in post 45
I get...

-1 + (3/2)(I(-4)(pi/2) - I(-4)(pi/4))
Where to from there?
 

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