Trigonometric Integration: Finding the Solution for ∫√(π) xsin(x²-1) dx

[Mutaja]

Homework Statement

₀∫^$\sqrt{∏}$ xsin($x^2$ -1) dx

Not sure how I should be formatting this, but the square root of pi is 'on top of' the integral, and zero is 'below'. The expression to integrate is $\sqrt{∏}$ xsin($x^2$ -1) dx.

The Attempt at a Solution

As sin integrated is -cos, I'm assuming that sin($x^2$ -1) integrated is -cos($x^2$ -1).

Using that logic, I get the following:

[-($\frac{1}{2}$x cos ($x^2$-1)) + c] (square root of pi on top, 0 below).

[-($\frac{1}{2}$ ∏ cos(∏-1))] + cos(-1) = 0,308 (using radians)

I feel that the solution is a bit weird, that I should be left out with a simpler expression or an answer like 0 or 1.

Am I integrating correctly? I fear not.

Thanks for any input on this problem in advance.

 

[LCKurtz]

Homework Statement

₀∫^$\sqrt{∏}$ xsin($x^2$ -1) dx

Not sure how I should be formatting this, but the square root of pi is 'on top of' the integral, and zero is 'below'. The expression to integrate is $\sqrt{∏}$ xsin($x^2$ -1) dx.

The Attempt at a Solution

As sin integrated is -cos, I'm assuming that sin($x^2$ -1) integrated is -cos($x^2$ -1).

Bad assumption. Does the derivative of $-\cos(x^2-1)$ give you back $\sin(x^2-1)$? Try a u substitution on your integral..

 

[Mutaja]

Bad assumption. Does the derivative of $-\cos(x^2-1)$ give you back $\sin(x^2-1)$? Try a u substitution on your integral..

I understand what you're saying, but I don't know how to do it. The way I understand the substitution method, I can only do it with two products, here I think I have 3. x, sin and $x^2$ -1.

If I use the substitution method, I want to rewrite $x^2$ -1 as u and get x sin u. What do I do next then? I remember doing something like f(x)g(x)h(x), but I'm not sure how that would work here.

Thanks for the idea, and I've obviously looked into it. I'm just not sure how to go on from here.

 

[FeDeX_LaTeX]

No, you only have a product of two things: $x$, and $\sin(x^2 - 1)$.

You're correct that you need to use the substitution $u = x^2 - 1$. What, then, is $\frac{du}{dx}$?

 

[Mutaja]

No, you only have a product of two things: $x$, and $\sin(x^2 - 1)$.

You're correct that you need to use the substitution $u = x^2 - 1$. What, then, is $\frac{du}{dx}$?

2x?

I'm not even close to following this logic in terms of integrating my original expression. How am I supposed to think when I look at my expression?

I'm both stuck and confused, unfortunately.

 

[Mark44]

I understand what you're saying, but I don't know how to do it. The way I understand the substitution method, I can only do it with two products, here I think I have 3. x, sin and $x^2$ -1.

NO! As already mentioned this product consists of two factors: x and sin(x² - 1).

It is vital that you understand that an expression such as sin(x² - 1) does NOT mean sin * (x² - 1), any more than log 6 means log * 6 or that √3 means √ * 3. I hope that you recognize that the three misinterpretations are meaningless. Both sin and log are functions, not numbers being multiplied.

If I use the substitution method, I want to rewrite $x^2$ -1 as u and get x sin u.

You're getting ahead of yourself. If u = x² - 1, then du/dx = 2x, or equivalently, du = 2x dx.

What do I do next then? I remember doing something like f(x)g(x)h(x), but I'm not sure how that would work here.

You should be thinking of the chain rule, which is really what ordinary substitution is about.

Thanks for the idea, and I've obviously looked into it. I'm just not sure how to go on from here.

2x?

I'm not even close to following this logic in terms of integrating my original expression. How am I supposed to think when I look at my expression?

I'm both stuck and confused, unfortunately.

 

[HallsofIvy]

First, forget the trig part- that is not what is causing you trouble! Suppose you had $$\int (x^2- 1)^5xdx$$. What do you get when you make the substitution $$u= x^2- 1$$?

 

[Mutaja]

First, forget the trig part- that is not what is causing you trouble! Suppose you had $$\int (x^2- 1)^5xdx$$. What do you get when you make the substitution $$u= x^2- 1$$?

As I'm on my phone, I can't reply to the above post for some reason. I'll get back to later.

To this post:

I get ∫$u^5$ = [itex]\frac{##u^6}{6}[/itex].

Sorry if I made a sloppy mistake here but it should be right.

Oh, I didn't notice the x there. It confuses me, but I'll have a look at it later using integration by parts.

Thanks for your help, I'm sorry I'm a little busy right now to do this properly. I'll get back to this in a few hours.

 

[Mark44]

As I'm on my phone, I can't reply to the above post for some reason. I'll get back to later.

To this post:

I get ∫$u^5$ = [itex]\frac{##u^6}{6}[/itex].

You have a mix of itex and ## tags that are causing the above to not render correctly. Here is the fixed version of what you wrote:
$∫u^5 = \frac{u^6}{6}$.

You are omitting something that is very important - the differential. The integral above should be ∫u⁵ du. You started off with something of this form: ∫f(x) dx. Using substitution, you need to replace the x expressions and dx with some expression in u plus du that is easier to integrate. If you ignore dx and du, you won't be able to perform the integration successfully.

Sorry if I made a sloppy mistake here but it should be right.

Oh, I didn't notice the x there. It confuses me, but I'll have a look at it later using integration by parts.

No, that's a bad move. Both integrals can and should be done using an ordinary substitution, which is one of the easier integration techniques, and one that should be tried before attempting the more complicated techniques (such as integration by parts).

Thanks for your help, I'm sorry I'm a little busy right now to do this properly. I'll get back to this in a few hours.

 

[Mutaja]

Ok, I've gone through this once again, and I was clearly rushing through it and I'm sorry for wasting your time. I can ensure you that this won't happen again.

I think I have a better understanding of it now, although I still encounter difficulties with substituting back. But I won't get ahead of myself this time, so let me show you what I've done, starting from scratch, using your guidelines.

Ignore the values for now, as I'm not sure how to format them correctly.

∫xsin($x^2$-1) dx

u = $x^2$-1

$\frac{du}{dx}$ = 2x -> du = 2x dx

Since du = 2x dx I'm assuming I can substitute x for $\frac{1}{2}$ du to get the following:

∫$\frac{1}{2}$ sinu du -> [$\frac{cosu^\frac{3}{2}}{3}$]

Now, let's take the 'limits' into play again. The upper limit is $\sqrt{∏}$, the lower limit is 0. After my last step, when the expression is integrated and put into [ ], I'm assuming that they're no longer the same. As I use substitution on the expression, I should do something with the value of the limits as well I think.

I will look into this, but feel free to look through my work now and see if it's better. At least it makes sense to me after taking what you've said into consideration, but I might've overlooked or ignored something.

Appreciate what you've done so far. Thanks.

 

[Mark44]

Ok, I've gone through this once again, and I was clearly rushing through it and I'm sorry for wasting your time. I can ensure you that this won't happen again.

I think I have a better understanding of it now, although I still encounter difficulties with substituting back. But I won't get ahead of myself this time, so let me show you what I've done, starting from scratch, using your guidelines.

Ignore the values for now, as I'm not sure how to format them correctly.

∫xsin($x^2$-1) dx

u = $x^2$-1

$\frac{du}{dx}$ = 2x -> du = 2x dx

Since du = 2x dx I'm assuming I can substitute x for $\frac{1}{2}$ du to get the following:

Since du = 2x dx, you can replace x dx with (1/2)du.

In your integral, ∫x sin(u) dx
becomes (1/2)∫sin(u)du.

∫$\frac{1}{2}$ sinu du -> [$\frac{cosu^\frac{3}{2}}{3}$]

No. You're trying to do too many things at one time. It's much simpler than that. What's an antiderivative of sin(u)? When you have that, replace u with x² - 1, and then evaluate at the two limits of integration.

Now, let's take the 'limits' into play again. The upper limit is $\sqrt{∏}$, the lower limit is 0. After my last step, when the expression is integrated and put into [ ], I'm assuming that they're no longer the same. As I use substitution on the expression, I should do something with the value of the limits as well I think.

I will look into this, but feel free to look through my work now and see if it's better. At least it makes sense to me after taking what you've said into consideration, but I might've overlooked or ignored something.

Appreciate what you've done so far. Thanks.

 

[Mutaja]

No. You're trying to do too many things at one time. It's much simpler than that. What's an antiderivative of sin(u)? When you have that, replace u with x² - 1, and then evaluate at the two limits of integration.

Ok, so I have $\frac{1}{2}$ ∫ sin(u) du.

The antiderivative of sin(u) is - cos(u).

I now have $\frac{1}{2}$ [ cos(u) ] with upper limit $\sqrt{∏}$, lower limit 0.

$\frac{1}{2}$ [ cos($x^2$-1) ]

Take the whole expression, put $\sqrt{∏}$ in for x, then subtract the whole expression with 0 for x from that.

$\frac{1}{2}$ [-cos(∏ - 1)] - [-cos(0-1]. How can I simplify this? And is it correct so far?

I get $\frac{1}{2}$ (cos(∏-1) + cos(-1) = 0.9996...

 

[LCKurtz]

You have dropped a minus sign but otherwise is is now correct. Can you find it? You should get about .54 for a answer.