Asked by Z32
Evaluate the integral from 2/sqrt2 to 4; 1/(t^3 sqrt(t^2-4))?
Answers
Answered by
drwls
The indefinite integral is
[1/(8t^2)]*sqrt(t^2-4) + (1/16)cos^-1(2/t)
I got it from a table of integrals.
Evaluate it at t = 4 and t = sqrt2
and take the difference.
Note that 2/sqrt2 = sqrt2.
There is a problem with your limits of integration. The function is undefined for t < 2, and yet you are trying to start the integration at sqrt2
[1/(8t^2)]*sqrt(t^2-4) + (1/16)cos^-1(2/t)
I got it from a table of integrals.
Evaluate it at t = 4 and t = sqrt2
and take the difference.
Note that 2/sqrt2 = sqrt2.
There is a problem with your limits of integration. The function is undefined for t < 2, and yet you are trying to start the integration at sqrt2
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