Chapter A6, Problem 41E

To determine

Toevaluate: the given integrals.

Answer to Problem 41E

The required result is

  ${\displaystyle \int \sinh }(2x)dx=\frac{\cosh 2x}{2}+C$

Explanation of Solution

Given:

  ${\displaystyle \int \sinh }(2x)dx$

Calculation:

The given integral is ${\displaystyle \int \sinh }(2x)dx$

assume $u=2x$

Therefore $du=2dx$

Now

  $\begin{array}{l}{\displaystyle \int \sinh }(2x)dx=\frac{1}{2}{\displaystyle \int \sinh }(u)\cdot du\\ =\frac{\cosh u}{2}+C\\ =\frac{\cosh 2x}{2}+C\\ \Rightarrow \frac{\cosh 2x}{2}+C\\ {\displaystyle \int \sinh }(2x)dx=\frac{\cosh 2x}{2}+C\end{array}$

Conclusion:

The required result is

  ${\displaystyle \int \sinh }(2x)dx=\frac{\cosh 2x}{2}+C$