Toronto Math Forum
MAT2442018F => MAT244Lectures & Home Assignments => Topic started by: Jiacheng Ge on November 06, 2018, 10:07:24 PM

Can anyone explain the integral here? I tried by parts but don't know how.

nvm

$$\frac{e^t}{2}\int{e^{t}\tanh{(t)}} dt$$
For the integrand ${e^{t}\tanh{t}}$, write $\tanh{t}$ as $\frac{e^te^{t}}{e^{t}+e^t}$.
(Note that if this doesn't look familiar, you should review hyperbolic sine and cosine)
$$\frac{e^t}{2}\int{e^{t}\tanh{(t)}}dt=\frac{e^t}{2}\int{\frac{e^{t}(e^te^{t})}{e^{t}+e^t}}dt$$
Expanding the integrand gives
$$=\frac{e^t}{2}\int{\left(\frac{e^t}{e^{2t}+1}\frac{e^{t}}{e^{2t}+1}\right)}dt=\frac{e^t}{2}\int{\left(\frac{e^t}{e^{2t}+1}\frac{e^{t}}{e^{2t}+1}\right)}dt$$
For the integrand $\left(\frac{e^t}{e^{2t}+1}\right)$, substitute $u=e^x$ and $du=e^x dx$, and for the integrand $\left(\frac{e^{t}}{e^{2t}+1}\right)$, substitute $v=e^x$ and $dv=e^x dx$.
Try proceeding this way. If you need help beyond this point, just lmk

Can you show me the detailed integral of y2?

It is not the easiest way
$$
\int e^{t}\tanh(t)\,dt = \int \frac{e^{t}e^{t}}{e^{t}+e^{t}}e^{t}\,dt= \int \frac{1e^{2t}}{1+e^{2t}}e^{t}\,d=
\int \Bigl( 1 \frac{2}{1+e^{2t}}\Bigr)\,de^{t} = e^{t}2\arctan (e^{t})+c.$$