7-7-51 integral of coth
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To evaluate this integral, we're going to split hyperbolic
cotangent into hyperbolic cosine divided by hyperbolic sine, and
we're going to do AU sub.
So if I let U equal the hyperbolic sine of 6X, then our
DU is 6 hyperbolic cosine 6 XDX or 16 DU equals hyperbolic
cosine 6 XDX.
So when we substitute here, that hyperbolic cosine 6 XDX just
gets replaced with 160 U, and that leaves me one over U.
If I go ahead and change my bounds, I get U to the LN5, so
that's going to equal the hyperbolic sine of six times
LN5.
Using my definition, I get E to the LN5 to the 6th minus E to
the LN5 to the -6 all over 2.
The E to the LNS cancels, so 5 to the 6th -1 / 5 to the 6th all
over two doing the bottom bound U of LN3.
Same exact skills, hyperbolic sine of six times Ln threes.
Take the six up in front.
Put it into the official definition.
So when I evaluate that integral, we get 1/6 the natural
log of the absolute value of U sticking in the upper bound,
then minus the lower bound.
That gives us approximately .5108 direction, say round to
the nearest hundredth, so .51.