Asked by Parker
"Leave the answer as a definite integral, but indicate how it could by evaluated by using the fundamental theorem of calculus."
I solved the problem to a definite integral. Proceeding via the fundamental theorem, would involve finding the indefinite integral (or antiderivative). I can use a computer program to do this, but I don't see a reasonable way to do this by hand. Any suggestions? thanks!
$\int_1^2 \frac{2\pi}{x} \sqrt{1 + \frac{1}{x^4}} \, \diff x$
I solved the problem to a definite integral. Proceeding via the fundamental theorem, would involve finding the indefinite integral (or antiderivative). I can use a computer program to do this, but I don't see a reasonable way to do this by hand. Any suggestions? thanks!
$\int_1^2 \frac{2\pi}{x} \sqrt{1 + \frac{1}{x^4}} \, \diff x$
Answers
Answered by
Count Iblis
Substituting x = sqrt(t) leads to an integral of the form:
Integral of dt/t sqrt(1+t^2)
If you then put t = sinh(u), this becomes:
Integral of cosh^2(u)/sinh(u)du =
Integral of [1/sinh(u) + sinh(u)] du
Then the integral of 1/sinh(u) be evaluated by putting u = Log(v):
du/[exp(u) - exp(u)] =
dv/[v (v - 1/v)] = dv/(v^2 - 1)
which is easily integrated.
Integral of dt/t sqrt(1+t^2)
If you then put t = sinh(u), this becomes:
Integral of cosh^2(u)/sinh(u)du =
Integral of [1/sinh(u) + sinh(u)] du
Then the integral of 1/sinh(u) be evaluated by putting u = Log(v):
du/[exp(u) - exp(u)] =
dv/[v (v - 1/v)] = dv/(v^2 - 1)
which is easily integrated.
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