Chapter 7.2, Problem 23E

To determine

To calculate:

To evaluate the given integral ${\displaystyle \int {\left(1-\cos \frac{t}{2}\right)}^2\sin \frac{t}{2}dt}$ with the initial condition $u=1-\cos \frac{t}{2}$

Answer to Problem 23E

The solution is $\frac{2}{3}{\left(1-\cos \frac{t}{2}\right)}^3+c$

Explanation of Solution

Given information:

The given integral is ${\displaystyle \int {\left(1-\cos \frac{t}{2}\right)}^2\sin \frac{t}{2}dt}$ with the initial condition $u=1-\cos \frac{t}{2}$

Consider the initial condition $u=1-\cos \frac{t}{2}$ ,

Differentiating both sides with respect to x,

  $du=0-\left(-\sin \frac{t}{2}\right)\cdot \frac{1}{2}\Rightarrow \frac{1}{2}\sin \frac{t}{2}$

Since the integral has $\sin \frac{t}{2}$ , solve this for $\sin \frac{t}{2}$ ,

  $\begin{array}{l}\frac{du}{dr}=\frac{1}{2}\sin \frac{t}{2}\\ du=\frac{1}{2}\sin \frac{t}{2}dt\\ 2du=\sin \frac{t}{2}dt\end{array}$

Substitute $u=1-\cos \frac{t}{2}$ and $2du=\sin \frac{t}{2}dt$ into the integral ${\displaystyle \int {\left(1-\cos \frac{t}{2}\right)}^2\sin \frac{t}{2}dt}$ ,

  ${\displaystyle \int {\left(1-\cos \frac{t}{2}\right)}^2\sin \frac{t}{2}dt}={\displaystyle \int u^2\left(2du\right)}$

  $=2{\displaystyle \int u^{2}du}$

On integrating we get,

  $\begin{array}{l}=2\left(\frac{u^3}{3}\right)+c\\ =\frac{2}{3}{u}^3+c\end{array}$

Now substituting back for u,

  $=\frac{2}{3}{\left(1-\cos \frac{t}{2}\right)}^3+c$

Therefore, the solution is $\frac{2}{3}{\left(1-\cos \frac{t}{2}\right)}^3+c$