Author Topic: quiz 5 0701  (Read 122 times)

Jiacheng Ge

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quiz 5 0701
« on: November 06, 2018, 10:07:24 PM »
Can anyone explain the integral here? I tried by parts but don't know how.
« Last Edit: November 06, 2018, 10:57:05 PM by Jiacheng Ge »

Jiacheng Ge

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Re: integral
« Reply #1 on: November 06, 2018, 10:56:49 PM »
 nvm
« Last Edit: November 06, 2018, 11:34:44 PM by Jiacheng Ge »

Monika Dydynski

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Re: integral
« Reply #2 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


Jiacheng Ge

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Re: quiz 5 0701
« Reply #3 on: November 06, 2018, 11:48:25 PM »
Can you show me the detailed integral of y2?

Victor Ivrii

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Re: quiz 5 0701
« Reply #4 on: November 07, 2018, 12:54:15 AM »
It is not the easiest way
$$
\int e^{-t}\tanh(t)\,dt = \int \frac{e^{t}-e^{-t}}{e^{t}+e^{-t}}e^{-t}\,dt= \int \frac{1-e^{-2t}}{1+e^{-2t}}e^{-t}\,d=
\int \Bigl(  1- \frac{2}{1+e^{-2t}}\Bigr)\,de^{-t} = e^{-t}-2\arctan (e^{-t})+c.$$