Chapter 6.3, Problem 17E

To determine

To find: The unknown integral.

Answer to Problem 17E

The unknown integral: $\frac{1}{2}{e}^x(\sin x-\cos x)+C$ . Here C is a constant.

Explanation of Solution

Given information:

The integration is: $\int e^x\sin xdx$ .

Calculation:

The given integration is: $\int e^x\sin xdx$ .

The formula for integration by parts $\int udv=uv-\int vdu$ To use LIPET (Logarithms, Inverse Trig, Polynomial, Exponential, Trig) $u$ should be $e^x$

So, $dv=\sin xdx$

(Both $e^x$ and $\sin x$ are cylindrical choose either to be $e^x$ )

Then,

  $\begin{array}{c}du=e^{x}dx\\ v=-\cos x\end{array}$

Also,

  $\begin{array}{c}\int e^x\sin xdx=e^x(-\cos x)-\int (-\cos x)\left(e^{x}dx\right)\\ =-e^x\cos x+\int e^x\cos xdx\end{array}$

Integration by parts must be employed once again because the integral on the right is still a product.

Let,

  $\begin{array}{c}u=e^x\\ dv=\cos xdx\end{array}$

Then,

  $\begin{array}{c}du=e^{x}dx\\ v=\sin x\end{array}$

  $\begin{array}{c}\int e^x\sin xdx=-e^x\cos x+\left[e^x\sin x-\int (\sin x)\left(e^{x}dx\right)\right]\\ \int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx\\ 2\int e^x\sin xdx=-e^x\cos x+e^x\sin x+C\\ \int e^x\sin xdx=-\frac{1}{2}{e}^x\cos x+\frac{1}{2}{e}^x\sin x+C\\ =\frac{1}{2}{e}^x(\sin x-\cos x)+C\end{array}$

Therefore, the required unknown integral of the given integration $\int e^x\sin xdx$ is $\frac{1}{2}{e}^x(\sin x-\cos x)+C$ . Here C is a constant.