Chapter 7.2, Problem 17E

To determine

To evaluate the integral, using the indicated substitution.

Answer to Problem 17E

  $\text{The required solution is, }-\frac{1}{3}\cos 3x+C.$

Explanation of Solution

Given information: $\text{The equation is},{\displaystyle \int \sin x}dx\text{ , }u=3x\text{.}$

Calculation : The required equation is obtained as,

  $\begin{array}{l}{\displaystyle \int \sin x}dx={\displaystyle \int \sin u}\left(\frac{1}{3}du\right)\left(\begin{array}{l}\text{If }u=3x,\text{ differentiating both side gives }du=3dx.\\ \text{solving this for }dx\text{ then gives }dx=\frac{1}{3}du.\text{ Substitute}\\ u=3x\text{ and }dx=\frac{1}{3}du\text{ into the integral}\text{.}\end{array}\right)\\ =\frac{1}{3}{\displaystyle \int \sin u}du\\ =\frac{1}{3}\left(-\cos u\right)+C\\ =-\frac{1}{3}\cos u+C\\ =-\frac{1}{3}\cos 3x+C\left(\text{substitute }u=3x\right)\\ \text{Hence, the required solution is, }-\frac{1}{3}\cos 3x+C.\end{array}$