Is this Tricky Definite Integral Solvable?

[pierce15]

1. The problem, the whole problem, and nothing but the problem

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx$$

Homework Equations

Integration by parts
trig substitution

The Attempt at a Solution

My first idea was to break up the integral by letting $u=x$ and $dv=sin(x)/(1+cos^2 x)$. I will omit the work, but it got me here:

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx = (x \cdot tan^{-1} (cos(x)) \big|_0^\pi - \int_0^\pi tan^{-1}(cos(x)) \, dx$$

I am fairly sure that I arrived at the second integral correctly. Is it integrateable, if that's a word?

 

[Dick]

1. The problem, the whole problem, and nothing but the problem

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx$$

Homework Equations

Integration by parts
trig substitution

The Attempt at a Solution

My first idea was to break up the integral by letting [inline]u=x[/inline] and [inline]dv=sin(x)/(1+cos^2 x)[/inline]. I will omit the work, but it got me here:

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2(x)} \, dx = (x \cdot tan^{-1} (cos(x)) \big|_0^\pi - \int_0^\pi tan^{-1}(cos(x)) \, dx$$

I am fairly sure that I arrived at the second integral correctly. Is it integrateable, if that's a word?

Also, what is the inline command? I forget it every time I go on this sight.

$works for inline. For the integral, it is a trick. Try substituting u=pi-x. Finding an indefinite integral is really pretty hopeless.$

 

[pierce15]

$$- \int_0^\pi tan^{-1} cos(x) \, dx$$
$$u= \pi-x, du=- dx$$
$$\int_\pi^0 tan^{-1} cos(\pi-u) \, du$$
$$- \int_0^\pi tan^{-1}(cos(\pi)cos(u) + sin(\pi)sin(u)) \, du$$
$$- \int_0^\pi tan^{-1}(-cos(u)) \, du$$
$$\int_0^\pi tan^{-1}cos(u) \, du$$

Hmm... does that mean that since the top equation is equal to the bottom, and they are negatives, the integral must be 0?

 

[Dick]

$$- \int_0^\pi tan^{-1} cos(x) \, dx$$
$$u= \pi-x, du=-x \, dx$$
$$\int_\pi^0 tan^{-1} sin(u) \, du$$
$$-\int_0^\pi tan^{-1} sin(u) \, du$$

OK, what now? Maybe:

$$\int_0^\pi tan^{-1} sin(u) \, du = \int_0^\pi tan^{-1} cos(x) \, dx$$

Is that what you had in mind?

I meant you to use that substitution in the original integral. You should get the original integral back with a different sign plus something you can integrate. Equate it to the original integral.

 

[pierce15]

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx$$
$$u=\pi-x, du=-1$$
$$- \int_\pi^0 \frac{(\pi-u)(sin(\pi-u))}{1+cos^2(\pi-u)} \, du$$
$$\int_0^\pi \frac{(\pi-u)(-sin(u))}{1+ cos^2(\pi-u)} \, du$$
$$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2(\pi-u)} \, du$$
$$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ (-cos(u))^2} \, du$$
$$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2u} \, du$$
$$-\pi \int_0^\pi \frac{sin(u)}{1+ cos^2u} \, du + \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$
$$v=cos(u), dv=-sin(u) \, du$$
$$\pi \int_1^{-1} \frac{dv}{1+v^2} + \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$
Going back to the beginning:

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx = -\pi \int_{-1}^1 \frac{dv}{1+v^2} + \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$

$$0=\int_{-1}^1 \frac{dv}{1+v^2}$$

Am I on the right track?

 

[Dick]

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx$$
$$u=\pi-x, du=-1$$
$$- \int_\pi^0 \frac{(\pi-u)(sin(\pi-u))}{1+cos^2(\pi-u)} \, du$$
$$\int_0^\pi \frac{(\pi-u)(-sin(u))}{1+ cos^2(\pi-u)} \, du$$
$$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2(\pi-u)} \, du$$
$$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ (-cos(u))^2} \, du$$
$$-\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2u} \, du$$

What now?

First fix an error. sin(pi-u)=sin(u). Then break it into two integrals.

 

[pierce15]

$$\int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx$$
$$u=\pi-x, du=-1$$
$$- \int_\pi^0 \frac{(\pi-u)(sin(\pi-u))}{1+cos^2(\pi-u)} \, du$$
$$\int_0^\pi \frac{(\pi-u)(sin(u))}{1+ cos^2(\pi-u)} \, du$$
$$\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2(\pi-u)} \, du$$
$$\int_0^\pi \frac{sin(u)(\pi-u)}{1+ (-cos(u))^2} \, du$$
$$\int_0^\pi \frac{sin(u)(\pi-u)}{1+ cos^2u} \, du$$
$$\pi \int_0^\pi \frac{sin(u)}{1+ cos^2u} \, du - \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$
$$v=cos(u), dv=-sin(u) \, du$$
$$\pi \int_1^{-1} \frac{dv}{1+v^2} - \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u} \, du$$

Is the following OK:

$$I = \int_0^\pi \frac{x \cdot sin(x)}{1+cos^2x} \, dx = \pi \int_{-1}^1 \frac{dv}{1+v^2} - \int_0^\pi \frac{u \cdot sin(u)}{1+ cos^2u}$$

$$2I = \pi \int_{-1}^1 \frac{dv}{1+v^2}$$

If so, how can I evaluate this integral?

 

[Dick]

Am I on the right track?

You know, having to constantly go back and check whether you have reedited a past post really sucks. Use preview to see whether you have what you want, then just LEAVE IT. If you change your mind, just post the correction. Don't go back and change the past. And yes, your result is fine.

 

[pierce15]

You know, having to constantly go back and check whether you have reedited a past post really sucks. Use preview to see whether you have what you want, then just LEAVE IT. If you change your mind, just post the correct. Don't go back and change the past.

Sorry

 

[Dick]

Sorry

S'ok. Just advice for the future. But you got the right answer.

 

[pierce15]

S'ok. Just advice for the future. But you got the right answer.

So the answer is just 0?

 

[Dick]

If so, how can I evaluate this integral?

It's an arctan, isn't it? Why do you think it's zero?

 

[Dick]

$$2I = \pi \int_{-1}^1 \frac{dv}{1+v^2}$$

If so, how can I evaluate this integral?

Now I forgot to quote the integral. Doesn't look like zero to me.

 

[pierce15]

It's an arctan, isn't it? Why do you think it's zero?

Oh, right. I'm pretty tired right now. Just did out the trig substitution and verified that it's 0. How did you know to make the substitution u=pi-x?

 

[pierce15]

$1/(1+v^2)$ is also an even function which is probably enough to show that the answer is 0, considering that the upper limit is equal to the lower limit

 

[Dick]

Oh, right. I'm pretty tired right now. Just did out the trig substitution and verified that it's 0. How did you know to make the substitution u=pi-x?

Tired, ok. But it's NOT zero. Check it in the morning if that works better. I guessed to do the substitution because I've seen problems like this before. Just experience.

 

[Dick]

$1/(1+v^2)$ is also an even function which is probably enough to show that the answer is 0, considering that the upper limit is equal to the lower limit

Wrong. Get some sleep if you need to.

 

[pierce15]

Tired, ok. But it's NOT zero. Check it in the morning if that works better. I guessed to do the substitution because I've seen problems like this before. Just experience.

$$\int_{-1}^1 \frac{dv}{v^2+1}$$
$$v= tan\theta, dv=sec^2\theta d\theta$$
$$\int_{-\pi/4}^{\pi/4} \frac{sec^2 \theta}{sec^2 \theta} d\theta$$
$$\int_{-\pi/4}^{\pi/4} \theta d\theta$$
$$\frac{\theta^2}{2} \big|_{-\pi/4}^{\pi/4}$$
$$(\pi/4)^2/2 - (-\pi/4)^2/2$$

Alright, I see that I'm wrong by looking at the graph. What's the problem here?

 

[Dick]

$$\int_{-1}^1 \frac{dv}{v^2+1}$$
$$v= tan\theta, dv=sec^2\theta d\theta$$
$$\int_{-\pi/4}^{\pi/4} \frac{sec^2 \theta}{sec^2 \theta} d\theta$$
$$\int_{-\pi/4}^{\pi/4} \theta d\theta$$
$$\frac{\theta^2}{2} \big|_{-\pi/4}^{\pi/4}$$
$$(\pi/4)^2/2 - (-\pi/4)^2/2$$

Alright, I see that I'm wrong by looking at the graph. What's the problem here?

Now you are making me sleepy. sec(θ)^2/sec(θ)^2 isn't θ. It's 1.

 

[pierce15]

$$\theta \big|_{-\pi/4}^{\pi/4} = \pi/2$$
$$2I= \pi/2$$
$$I = \pi/4$$

Were there any arithmetic errors in there? haha

 

[Dick]

$$\theta \big|_{-\pi/4}^{\pi/4} = \pi/2$$
$$2I= \pi/2$$
$$I = \pi/4$$

Were there any arithmetic errors in there? haha

Sadly, yes. You dropped the factor of pi you had correct in post 7. You would do it easily with the hint I gave you if you were less tired.

 

[pierce15]

Alright, so $I = \pi^2/4$. Thank you very much for the help, now I'm going to get some beauty sleep

 

[Dick]

Alright, so $I = \pi^2/4$. Thank you very much for the help, now I'm going to get some beauty sleep

Very welcome, and good idea!