Chapter 5.6, Problem 14E

To determine

Tofind: the value of integral.

Answer to Problem 14E

The value of integral is $I=\frac{1}{5}{e}^{-\theta }\left(2\sin 2\theta -\cos 2\theta \right)+c$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle \int e^{-\theta }\cos 2\theta d\theta }$ .

Formula used: integral by parts,

  ${\displaystyle \int u·vdx=u{\displaystyle \int vdx}}-{\displaystyle \int u\text{'}}{\displaystyle \int vdx}dx$

Calculation: to find the value of integral, first use the integral by parts formula then simplify it,

  $\begin{array}{l}I={\displaystyle \int e^{-\theta }\cos 2\theta d\theta }\\ I=\cos 2\theta {\displaystyle \int e^{-\theta }d\theta }-{\displaystyle \int \left(-2\sin 2\theta \right)}\left(\frac{e^{-\theta }}{-1}\right)d\theta \text{ [use integral by parts and simplify]}\\ I=-e^{-\theta }\cos 2\theta -2{\displaystyle \int \sin 2\theta e^{-\theta }d\theta }\text{ [integrate and simplify]}\\ I=-e^{-\theta }\cos 2\theta -2\left[-e^{-\theta }\sin 2\theta -{\displaystyle \int \left(-2\cos \theta e^{-\theta }\right)}\right]\text{ [again use integral by parts and integrate]}\\ I=-e^{-\theta }\cos 2\theta +2e^{-\theta }\sin 2\theta -4I\text{ [multiply and simplify]}\\ I+4I=-e^{-\theta }\cos 2\theta +2e^{-\theta }\sin 2\theta \text{ [add]}\\ I=\frac{1}{5}{e}^{-\theta }\left(2\sin 2\theta -\cos 2\theta \right)+c\text{ [add , divide and simplify]}\end{array}$

Thus, the value of integral is $I=\frac{1}{5}{e}^{-\theta }\left(2\sin 2\theta -\cos 2\theta \right)+c$ .