Hi, I hope someone can help with this problem:
Show that
integral_{x}^{1} dt/ (1+ t^2)
is equal to
integral_{1}^{1/x} dt/ (1 + t^2) if x>0.
Thanks
I'm assuming that _{x}^{1} means that the limits of integration are from x to 1, and that _{1}^{1/x} means that the limits of integration are from 1 to 1/x. If this is correct, I will rewrite the problem using my own notation.
INT{x,1} 1/(1 + t^2) dt = INT{1,1/x} 1/(1 + t^2) dt
It can be shown that:
INT 1/(1 + t^2) dt = arctan(t) + C
(If you need me to show how I know this, I can.)
Therefore, the above integrations become:
arctan(t) from {x,1} = arctan(t) from {1,1/x}
arctan(1) - arctan(x) = arctan(1/x) - arctan(1)
pi/4 - arctan(x) = arctan(1/x) - pi/4
Let arctan(x) = y --> tan(y) = x
Let arctan(1/x) = z --> tan(z) = 1/x --> cot(z) = x
Thus x = tan(y) = cot(z) --> y = pi/2 - z
So the above problem becomes:
pi/4 - y = z - pi/4
pi/4 - (pi/2 - z) = z - pi/4
-pi/4 + z = z - pi/4