How to solve the integral using integration by parts?

[t_n_p]

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

 

[HallsofIvy]

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.

 

[Gib Z]

Dido what Halls said, with tiny corrections:

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

 

[genneth]

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.

 

[HallsofIvy]

I was thinking right- my fingers hit the wrong keys!

(and it is "ditto", not "dido"!)

 

[t_n_p]

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?

 

[Gib Z]

Use the chain rule?

 

[t_n_p]

Let u = (n-1) hence y=[sin(x)]^(u)
then du/dn = 1

but again, unsure how to get dy/du now..

 

[Gib Z]

Instead, try letting u = sin x when differentiating :)

 

[genneth]

\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.

 

[t_n_p]

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)?

 

[t_n_p]

\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?

 

[genneth]

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...

 

[Gib Z]

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..

 

[t_n_p]

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?

 

[Gib Z]

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 :)

 

[t_n_p]

ah ok!
After converting (cosx)^2 using the trig identity I get..
http://img521.imageshack.us/img521/7374/untitledfv0.png
Where to from there?

 

[learningphysics]

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

 

[t_n_p]

ah ok, you mean expand to give..
http://img512.imageshack.us/img512/6858/untitleddl8.png

but then what?

 

[learningphysics]

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}

 

[EnumaElish]

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.

 

[t_n_p]

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...

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

 

[learningphysics]

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...

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...

 

[t_n_p]

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?

 

[learningphysics]

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.

 

[t_n_p]

After expanding and taking all a's to one side and dividing by n I get...

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

 

[learningphysics]

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.

 

[t_n_p]

my bad, crappy typo.

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

 

[learningphysics]

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).

 

[t_n_p]

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