Question

\[\int\limits_{}^{} 4\cosh(3\theta- \ln2)d \theta \]

Answer 1

This is a pretty simple \(u = 3\theta -\ln2\) substitution.

Answer 2

You have to consider that\[\cosh=\frac{ e^x+e^-x }{ 2 }\]\[\int\limits_{}^{}[4\cosh(3\theta-\ln(2))]d \theta\]Use u substitution \[u=3\theta-\ln(2)\]\[du=3d\theta\]\[d \theta=\frac{ du }{ 3 }\]Substitute u\[\int\limits_{}^{}4\cosh(u)\frac{ du }{ 3 }\]\[\frac{ 4 }{ 3 }\int\limits_{}^{}\cosh(u)du\]\[\frac{ 4 }{ 3 }\int\limits_{}^{}\frac{ e^u+e^-u }{ 2 }du\]\[\frac{ 4 }{ 3(2) }\int\limits_{}^{} e^u+(e^-u)du\]\[\frac{ 4 }{ 6 }[e^u+\ln|e^u|]\]Ten substitute the value of u\[u=3\theta-\ln(2)\]

Answer 3

\[\frac{ 4 }{ 6 }[e ^{3\theta-\ln2} + \ln|e ^{3\theta-\ln2}|] \] is this correct ? and also thank you for your help

Answer 4

yep that's right ^_^