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Recreational Problems

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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.  8)

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

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