`int_0^1 t cosh(t) dt` Evaluate the integral

Take the indefinite integral by parts. It is known (and easy to prove) that `cosh(t)=d/dt sinh(t)` and `sinh(t)= d/dt cosh(t).`

Denote `u=t,` `v=sinh(t),` then `du=dt` and `dv=cosh(t)dt.`

`int t cosh(t) dt= int u dv =uv- int v du=`

`=t sinh(t)-int sinh(t) dt= t sinh(t)-cosh(t)+C.`

Therefore the definite integral is

`sinh(1)-cosh(1)+1=1/2(e^1-e^(-1)-e^1-e^(-1))+1=1-1/e.`