Find the area of the region for ytanh6x from x0 to 2Solution

Find the area of the region for y=tanh(6x) from x=0 to 2

tanh(u)du

Rewrite using trigonometric/hyperbolic identities:

=sinh(u)cosh(u)du

Substitute v=cosh(u) dvdu=sinh(u)

=1vdv

This is a standard integral:

=ln(v)

Undo substitution v=cosh(u):

=ln(cosh(u))

hence tanh(u)du = ln(cosh(u))

Now Area = tanh(6x)dx = 1/6 (ln(cosh(6x))) and the limits for x are 0 to 2

Hence Area = 1/6 (ln(cosh(12)))=1.8844754699