Chapter A6, Problem 44E

To determine

Toevaluate: the given integrals.

Answer to Problem 44E

The required result is

  ${\displaystyle \int 4}\cosh (3x-\ln 2)\cdot dx=\frac{4}{3}\sinh (3x-\ln 2)+C$

Explanation of Solution

Given:

  ${\displaystyle \int 4}\cosh (3x-\ln 2)\cdot dx$

Calculation:

assume $u=3x-\ln 2$

Therefore $du=3dx$

Now

  $\begin{array}{l}{\displaystyle \int 4}\cosh (3x-\ln 2)dx=\frac{4}{3}{\displaystyle \int \cosh }(u)\cdot du\\ =\frac{4}{3}\sinh u+C\\ =\frac{4}{3}\sinh (3x-\ln 2)+C\\ {\displaystyle \int 4}\cosh (3x-\ln 2)\cdot dx=\frac{4}{3}\sinh (3x-\ln 2)+C\end{array}$

Conclusion:

The required result is

  ${\displaystyle \int 4}\cosh (3x-\ln 2)\cdot dx=\frac{4}{3}\sinh (3x-\ln 2)+C$