Chapter 6, Problem 21RE

To determine

To find: The value of integral.

Answer to Problem 21RE

The answer is: $\left(\frac{3\sin x}{10}-\frac{\cos x}{10}\right)e^{3x}+C$

Explanation of Solution

Given information:

The integration is: $\int e^{3x}\sin xdx$

Calculation:

Both ${\text{e}}^{3x}$ and $\sin x$ are cyclical so use tabular integration by parts with $f(x)=e^{3x}$ and

  $g(x)=\sin x$ Stop when a row has ${\text{e}}^{3x}$ and $\sin x$ . Then, beginning with $+$ , write alternate signs for each row. Draw an arrow from the left column to the expression one row lower in the right column for all rows except the last. Draw a straight line from the right column to the left column for the final row.

  

Using the symbols above each arrow, write out the results of the functions connected by the arrows. Since the final product wasn't a constant, it must be expressed as an integral. Don't forget to put $+C$ since its an indefinite integral.

  $\begin{array}{c}\int e^{3x}\sin xdx=e^{3x}(-\cos x)-3e^{3x}(-\sin x)+\int 9e^{3x}(-\sin x)dx\\ \to \int e^{3x}\sin xdx=-e^{3x}\cos x+3e^{3x}\sin x-9\int e^{3x}\sin xdx+C\\ \to 10\int e^{3x}\sin xdx=-e^{3x}\cos x+3e^{3x}\sin x+C\\ \to \int e^{3x}\sin xdx=-\frac{e^{3x}\cos x}{10}+\frac{3e^{3x}\sin x}{10}+C\\ =\left(\frac{3\sin x}{10}-\frac{\cos x}{10}\right)e^{3x}+C\end{array}$

Therefore the required integral of the given integration $\int e^{3x}\sin xdx$ is

  $\left(\frac{3\sin x}{10}-\frac{\cos x}{10}\right)e^{3x}+C$