How to solve the integral using integration by parts?

[t_n_p]

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?

 

[learningphysics]

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 $$\frac{\sqrt{2}}{4}$$

 

[t_n_p]

isnt that that first term to the power (n-1) making the second term to the poewr of -3?

 

[learningphysics]

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.

 

[t_n_p]

It was a typo, even you pointed that out :P

 

[learningphysics]

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) ??

 

[t_n_p]

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.

 

[learningphysics]

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...

 

[t_n_p]

yeah my bad, hence the lhs integral = -1
Now I solve the lhs integral?
Do I let n=0 and do it all again?

 

[learningphysics]

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.

 

[t_n_p]

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?

 

[learningphysics]

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...

 

[t_n_p]

Damn this is a difficult question!

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

 

[learningphysics]

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...

 

[t_n_p]

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?

 

[learningphysics]

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?

Yeah, what you've written is right. I've fixed the integrals. It's from pi/4 to pi/2 as you've written.

 

[t_n_p]

Ok, so once that is done I change back to the integral form?

 

[learningphysics]

Ok, so once that is done I change back to the integral form?

yeah. You can also go straight to the integral form after substituting in n=-2... it's up to you...

The idea is to get $$\int_{\pi/4}^{\pi/2}{sin^{-4}(x)dx}$$ in terms of $$\int_{\pi/4}^{\pi/2}{sin^{-2}(x)dx}$$

 

[t_n_p]

yeh, so once I get up to.. https://www.physicsforums.com/latex_images/14/1408407-0.png

Do I let n=0 and repeat process to solve for LHS?

 

[learningphysics]

yeh, so once I get up to.. https://www.physicsforums.com/latex_images/14/1408407-0.png

Do I let n=0 and repeat process to solve for LHS?

No. The left side can be solved directly... the integral of 1/sin^2(x) = -cot(x)...

 

[t_n_p]

oh, didnt think of that!

 

[learningphysics]

Did you finish the problem?

 

[t_n_p]

Yep! Thanks a bunch :D