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

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- April 30th 2007, 11:22 PMWintaker99integral proof
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 - May 1st 2007, 12:08 AMecMathGeek
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