Chapter 7, Problem 21RE

To determine

To evaluate the integral.

Answer to Problem 21RE

The required solution is, ${\displaystyle \int e^{3x}\sin xdx}=\left(\frac{3\sin x}{10}-\frac{\cos x}{10}\right)e^{3x}+C$

Explanation of Solution

Given information:The equation is, ${\displaystyle \int e^{3x}\sin xdx}$

Concept used: Tabular integration.

We have seen that integrals of the form ${\displaystyle \int f\left(x\right)}g\left(x\right)dx$ in which f can be differentiated repeatedly to become zero and g can be integrated repeatedly without difficulty, are natural candidates for integration by parts. However, if many repetitions are required, the calculation

Can be cumbersome.

Calculation : Using Tabular intergration, the required solution is obtained as,

with $f\left(x\right)=e^{3x}$ and $g\left(x\right)=\sin x$ , we list

  

We combine the products of the functions connected by the arrows according to the

Operation signs above the arrows to obtain,

  $\begin{array}{l}{\displaystyle \int e^{3x}\sin xdx}=e^{3x}\left(-\cos x\right)-3e^{3x}\left(-\sin x\right)+{\displaystyle \int 9e^{3x}\left(-\sin x\right)dx}\\ {\displaystyle \int e^{3x}\sin xdx}=-e^{3x}\cos x+3e^{3x}\sin x-9{\displaystyle \int e^{3x}\left(-\sin x\right)dx}+C\\ 10{\displaystyle \int e^{3x}\sin xdx}=-e^{3x}\cos x+3e^{3x}\sin x+C\\ {\displaystyle \int e^{3x}\sin xdx}=-\frac{e^{3x}\cos x}{10}+\frac{3e^{3x}\sin x}{10}+C\left(\text{Dividing both side by 10}\right)\\ {\displaystyle \int e^{3x}\sin xdx}=\left(\frac{3\sin x}{10}-\frac{\cos x}{10}\right)e^{3x}+C\end{array}$

Hence, the required solution is ${\displaystyle \int e^{3x}\sin xdx}=\left(\frac{3\sin x}{10}-\frac{\cos x}{10}\right)e^{3x}+C$