Chapter A6, Problem 43E

To determine

Toevaluate: the given integrals.

Answer to Problem 43E

The required result is

  ${\displaystyle \int 6}\cosh \left(\frac{x}{2}-\ln 3\right)\cdot dx=12\sinh \left(\frac{x}{2}-\ln 3\right)+C$

Explanation of Solution

Given:

  ${\displaystyle \int 6}\cosh \left(\frac{x}{2}-\ln 3\right)\cdot dx$

Calculation:

Assume $u=\frac{x}{2}-\ln 3$

Therefore $du=\frac{dx}{2}$

Now

  $\begin{array}{l}{\displaystyle \int 6}\cosh \left(\frac{x}{2}-\ln 3\right)dx=12{\displaystyle \int \cosh }(u)\cdot du\\ =12\sinh u+C\\ =12\sinh \left(\frac{x}{2}-\ln 3\right)+C\\ {\displaystyle \int 6}\cosh \left(\frac{x}{2}-\ln 3\right)\cdot dx=12\sinh \left(\frac{x}{2}-\ln 3\right)+C\end{array}$

Conclusion:

The required result is

  ${\displaystyle \int 6}\cosh \left(\frac{x}{2}-\ln 3\right)\cdot dx=12\sinh \left(\frac{x}{2}-\ln 3\right)+C$