Chapter 7.2, Problem 68E

(a)

To determine

To determine the integral ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}$ by substitution.

Answer to Problem 68E

The value of the integral ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}$ is $\frac{1}{2}$

Explanation of Solution

Given information: ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}$

Formula used: ${\displaystyle \int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$

Calculation :

Put $1-\cos 3x=t$ so that $3\sin 3xdx=dt$

Also, when $x=\pi /6;t=1$ and when $x=\pi /3;t=2$

The given integral becomes as:

  ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}=\frac{1}{3}{\displaystyle \underset{1}{\overset{2}{\int }}tdt}=\frac{1}{3}{\left[\frac{t^2}{2}\right]}_1^2=\frac{1}{6}\left[4-1\right]=\frac{1}{2}$

Hence, ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}=\frac{1}{2}$

(b)

To determine

To determine the integral ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}$ .

Answer to Problem 68E

The value of the integral ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}$ is $\frac{1}{2}$

Explanation of Solution

Given information: ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}$

Formula used: ${\displaystyle \int f^{\prime }(x).{\left[f(x)\right]}^{n}dx=\frac{{\left[f(x)\right]}^{n+1}}{n+1}+C}$

Calculation :

The given integral becomes as:

  ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}=\frac{1}{3}{\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}3(1-\cos 3x)\sin 3xdx}$

  $=\frac{1}{3}{\left[\frac{{(1-\cos 3x)}^2}{2}\right]}_{\pi /6}^{\pi /3}=\frac{1}{6}\left[4-1\right]=\frac{1}{2}$

Hence, ${\displaystyle \underset{\pi /6}{\overset{\pi /3}{\int }}(1-\cos 3x)\sin 3xdx}=\frac{1}{2}$