Recreational Problems

[Despondent]

It was more of a hint to others as to how to proceed rather than a personal attempt to get the answer out. One thing I observed whilst actually writing this one out was that over use of properties of the integrand can lead to dead ends.

The integral is
 
I = int(-pi,pi) (2x+2x*sin(x))/(1+(cos(x))^2) dx

= int(-pi,pi) ((2x)/(1+(cos(x))^2))dx + int(-pi,pi) (2x*sin(x))/(1+(cos(x))^2)dx

The first integral is zero because the integrand is odd and so

I = int(-pi,pi) (2x*sin(x))/(1+(cos(x))^2) dx

Rewriting the integral as 2 times an integral from 0 to pi here will actually end up making things more complicated.

= int(-pi,pi) (2x)*((sin(x))/(1+(cos(x))^2)dx

int ((sin(x))/(1+(cos(x))^2) dx = -arctan(cos(x)) + c

=> I = [-2x*arctan(cos(x))]|(-pi,pi) + 2*int(-pi,pi)(arctan(cos(x))) dx

= -2[x*arctan(cos(x))] |(-pi,pi) + 0 since the second integrand is odd

= -2(-(((pi)^2)/4)-(((pi)^2)/4))

So I = (pi)^2.

[Ahmad]

Can you justify why:
2*int(-pi,pi)(arctan(cos(x))) dx = 0, you say the integrand is odd, but it isn't.

I'll get more problems up when I have time.  

[Despondent]

If you've still got some exams to do it's probably best that you concentrate on those for now. Extra questions can come at any time.

Anyway, arctan(x) is an odd function. As long as its argument is continuous, any integral over arctan from (-b,b) (with b a real number) will always be zero.

Edit: Let me check.

[Ahmad]

Yes, but for a function to be odd:

f(-x) = -f(x)

Here you have f(x) = arctan[cos(x)]

But, arctan[cos(-x)] isn't equal to -arctan[cos(x)].

You have to use a special argument!  :wink:

[Despondent]

int(-pi,pi) arctan(cos(x)) dx

= 2*int(0,pi) arctan(cos(x)) dx

= 2*int(0,pi/2) arctan(cos(x)) dx + 2*int(pi/2,pi) arctan(cos(x)) dx

Let u = pi - x in the first integral then

arctan(cos(x)) = arctan(cos(pi-u)) = arctan(-cos(u))=-arctan(cos(u)) and so

I = 2*int(pi,pi/2) arctan(cos(u)) du + 2*int(pi/2,pi) arctan(cos(x)) dx

= -2*int*(pi/2,pi) arctan(cos(u)) du + 2*int(pi/2,pi) arctan(cos(x)) dx

= 0

[Ahmad]

If you need a hint, do ask.  It's really great how far you've got already, really well done!

[Despondent]

I should hardly need a hint considering that I did this sort of thing two years ago. Periodicity, it took a while to kick in. I'm a bit rusty.

[Ahmad]

That's correct. Well done. My solution is a little different, but it doesn't matter, they're both good.

Two years ago? You did this in year 10? Good stuff!  :wink:

[Despondent]

I'm in 3rd year uni.  My original intention was to just suggest a way to start this problem since there weren't any attempts. I didn't plan on doing the whole thing.

[Ahmad]

Cool. What are you taking in university?

[Despondent]

I'm taking BE/BSc.

Which course do you plan on taking?

[Ahmad]

Actuarial studies, so that's BComm I think. Although maths is my hobby/passion!  

[Despondent]

Well I hope you have fun taking it. Not sure if you know this already, but after first year a lot of the 'maths' that you would be doing will be prob/stats.

I've always disliked stats but I guess some people would enjoy it.