# Toronto Math Forum

## MAT244--2018F => MAT244--Lectures & Home Assignments => Topic started by: Jiacheng Ge on November 06, 2018, 10:07:24 PM

Title: quiz 5 0701
Post 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.
Title: Re: integral
Post by: Jiacheng Ge on November 06, 2018, 10:56:49 PM
nvm
Title: Re: integral
Post by: Monika Dydynski on November 06, 2018, 11:04:01 PM
$$-\frac{e^t}{2}\int{e^{-t}\tanh{(t)}} dt$$

For the integrand ${e^{-t}\tanh{t}}$, write $\tanh{t}$ as $\frac{e^t-e^{-t}}{e^{-t}+e^t}$.

(Note that if this doesn't look familiar, you should review hyperbolic sine and cosine)

$$-\frac{e^t}{2}\int{e^{-t}\tanh{(t)}}dt=-\frac{e^t}{2}\int{\frac{e^{-t}(e^t-e^{-t})}{e^{-t}+e^t}}dt$$

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

Title: Re: quiz 5 0701
Post by: Jiacheng Ge on November 06, 2018, 11:48:25 PM
Can you show me the detailed integral of y2?
Title: Re: quiz 5 0701
Post by: Victor Ivrii on November 07, 2018, 12:54:15 AM
It is not the easiest way
$$\int e^{-t}\tanh(t)\,dt = \int \frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}e^{-t}\,dt= \int \frac{1-e^{-2t}}{1+e^{-2t}}e^{-t}\,d= \int \Bigl( 1- \frac{2}{1+e^{-2t}}\Bigr)\,de^{-t} = e^{-t}-2\arctan (e^{-t})+c.$$