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Integrate: $\int^1_0\ t\cosh(t) dt$

I am little confused when it comes to problems such as this. Do I treat this values as inverses of cosine and sine respectively. Or do I utilize these formulas when I am about to evaluate after integrating?

Where $\sinh(x)= \frac{e^x-e^{-x}}{2}$ and $\cosh(x)= \frac{e^x+e^{-x}}{2}$.

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  • $\begingroup$ No $\pi$. We have $\cosh 0=1$ and $\cosh 1=\frac{e+e^{-1}}{2}$. $\endgroup$ Dec 2 '15 at 20:32
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$\begin{align} \int^1_0\ t\cosh(t) dt &= [t\sinh t]_0^1- \int_0^1 \sinh t dt \\ &= \sinh(1)- \cosh(1)+\cosh(0) \\ &= \frac{e-e^{-1}}{2} - \frac{e+e^{-1}}{2} +1 \\ &= -\frac{1}{e}+1 \\ &= \frac{e-1}{e} \end{align}$

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  • $\begingroup$ Can you explain your second line? $\endgroup$
    – Sunny
    Dec 2 '15 at 20:44
  • $\begingroup$ $\sinh(0)=0$ and by parts $\endgroup$
    – Trajan
    Dec 2 '15 at 20:44
  • $\begingroup$ sinh is the inverse of sin? so sinh(-1)=$\frac{3\pi}{2}$ $\endgroup$
    – Sunny
    Dec 2 '15 at 20:46
  • $\begingroup$ $-1$ doesn't come into it, the limits are $0$ and $1$. Only think of $\sinh$ in terms of exponentials. Rarely do you use the correspondence between $\sinh$ and $\sin$ $\endgroup$
    – Trajan
    Dec 2 '15 at 20:48
  • $\begingroup$ so $sinh(0)=0,\ sinh(1)=\sinh(x)= \frac{e^x-e^{-x}}{2},\ cosh(0)=1$ and $\cosh(x)= \frac{e^x+e^{-x}}{2}$ $\endgroup$
    – Sunny
    Dec 2 '15 at 20:51
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Notice, $$\int_{0}^{1}t\cosh(t)\ dt=\int_{0}^{1}t\left(\frac{e^t+e^{-t}}{2}\right)\ dt$$ $$=\left[t\left(\frac{e^t-e^{-t}}{2}\right)\right]_{0}^{1}-\left[\left(\frac{e^t+e^{-t}}{2}\right)\right]_{0}^{1}$$ $$=\left[\left(\frac{e-e^{-1}}{2}\right)-0\right]-\left[\frac{e+e^{-1}}{2}-1\right]$$ $$=\frac{e-e^{-1}-e-e^{-1}}{2}+1$$ $$=1-\frac{1}{e}=\color{red}{\frac{e-1}{e}}$$

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  • $\begingroup$ @sandstone: thanks for observation. I fixed the error $\endgroup$ Dec 2 '15 at 21:03
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$$\int_{0}^{1}t\cosh(t)\space\text{d}t=$$


For the intergand $t\cosh(t)$, integrate by parts: $\int f\space\text{d}g=fg-\int g\space\text{d}f$ where

$f=t\space,\space\text{d}g=\cosh(t)\space\text{d}t\space,\space\text{d}f=\text{d}t\space,\space g=\sinh(t)$:


$$\left[t\sinh(t)\right]_{0}^{1}-\int_{0}^{1}\sinh(t)\space\text{d}t=$$ $$\left[t\sinh(t)\right]_{0}^{1}-\left[\cosh(t)\right]_{0}^{1}=$$ $$\left(1\sinh(1)-0\sinh(0)\right)-\left(\cosh(1)-\cosh(0)\right)=$$ $$\left(-\frac{1-e^2}{2e}-0\right)-\left(\frac{1+e^2}{2e}-1\right)=$$ $$\left(-\frac{1-e^2}{2e}\right)-\left(\frac{1+e^2}{2e}-1\right)=$$ $$-\frac{1-e^2}{2e}-\frac{1+e^2}{2e}+1=\frac{e-1}{e}\approx 0.63212$$

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