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December 21, 2014

December 21, 2014

Posted by **Heather** on Tuesday, November 13, 2012 at 6:01pm.

- Calculus 2 -
**Steve**, Tuesday, November 13, 2012 at 6:10pmif x = pi*tan(z)

pi^2 + x^2 = pi^2 + pi^2 tan^2(z) = pi^2 sec^2(z)

dx = pi*sec^2(z)

1/(pi^3 sec^3(z)) * pi*sec^2(z) dz

= 1/pi^2 cos(z) dz

the integral is thus 1/pi^2 sin(z)

x/pi * tan(z), so

sin(z) = x/sqrt(pi^2 + x^2)

the integral is thus x/[pi^2 sqrt(pi^2 + x^2)]

from 0 to infinity is thus 1/pi^2

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