Chapter 6.2, Problem 60E

To determine

To evaluate: the given integral.

Answer to Problem 60E

  ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}{\cos }^{-3}2\theta \sin 2\theta d\theta }=\frac{3}{4}$ .

Explanation of Solution

Given:

  ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}{\cos }^{-3}2\theta \sin 2\theta d\theta }$ .

Concept used:

The Fundamental Theorem of Calculus, Evaluation Part:

If $f$ is continuous at every point of $\left[a,b\right]$ , and if $F$ is any antiderivative of $f$ on $\left[a,b\right]$ , then

  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=F\left(b\right)-F\left(a\right)$ .

Calculation:

Consider the given integral,

  ${\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}{\cos }^{-3}2\theta \sin 2\theta d\theta }$

Put $u=\cos 2\theta$ . Then,

  $\begin{array}{c}\frac{du}{d\theta }=-2\sin 2\theta \\ -\frac{du}{2}=\sin 2\theta d\theta \end{array}$

Now,

  $\begin{array}{c}u\left(0\right)=\cos 2\left(0\right)\\ =\cos 0\\ =1\end{array}$

And,

  $\begin{array}{c}u\left(\frac{\pi }{6}\right)=\cos 2\left(\frac{\pi }{6}\right)\\ =\cos \frac{\pi }{3}\\ =\frac{1}{2}\end{array}$

Therefore the integral becomes,

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}{\cos }^{-3}2\theta \sin 2\theta d\theta }={\displaystyle \underset{1}{\overset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}{u}^{-3}\left(-\frac{du}{2}\right)}\\ =-\frac{1}{2}{\displaystyle \underset{1}{\overset{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}{u}^{-3}du}\\ =-\frac{1}{2}{\left[\frac{u^{-3+1}}{-3+1}\right]}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =-\frac{1}{2}{\left[\frac{u^{-2}}{-2}\right]}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ =\frac{1}{4}{\left[\frac{1}{u^2}\right]}_1^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}$

Now by applying the limits, it follows that

  $\begin{array}{c}{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\int }}{\cos }^{-3}2\theta \sin 2\theta d\theta }=\frac{1}{4}\left[\frac{1}{{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2}-\frac{1}{1^2}\right]\\ =\frac{1}{4}\left[4-1\right]\\ =\frac{3}{4}\end{array}$