Integrate: $\int^1_0\ t\cosh(t)\ dt$ 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 $\cosh(x)= \frac{e^x+e^{-x}}{2}$.

 A: $\begin{align} \int^1_0\ t\cosh(t) dt &= [t\sinh t]_0^1- \int_0^1 \sinh t dt \\
&= \sinh(1)- \cosh(1)+\cosh(0) \\
&= \frac{e-e^{-1}}{2} - \frac{e+e^{-1}}{2} +1 \\
&= -\frac{1}{e}+1 \\
&= \frac{e-1}{e}
\end{align}$
A: Notice, $$\int_{0}^{1}t\cosh(t)\ dt=\int_{0}^{1}t\left(\frac{e^t+e^{-t}}{2}\right)\ dt$$
$$=\left[t\left(\frac{e^t-e^{-t}}{2}\right)\right]_{0}^{1}-\left[\left(\frac{e^t+e^{-t}}{2}\right)\right]_{0}^{1}$$
$$=\left[\left(\frac{e-e^{-1}}{2}\right)-0\right]-\left[\frac{e+e^{-1}}{2}-1\right]$$ $$=\frac{e-e^{-1}-e-e^{-1}}{2}+1$$  $$=1-\frac{1}{e}=\color{red}{\frac{e-1}{e}}$$
A: $$\int_{0}^{1}t\cosh(t)\space\text{d}t=$$

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

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