Chapter 5.5, Problem 41E

To determine

To evaluate: The definite integral.

Answer to Problem 41E

The integration of the definite integral ${\displaystyle \underset{0}{\overset{1}{\int }}\cos \left(\frac{\pi t}{2}\right)}dt$ is $\frac{2}{\pi }$ .

Explanation of Solution

Given information: ${\displaystyle \underset{0}{\overset{1}{\int }}\cos \left(\frac{\pi t}{2}\right)}dt$

Calculation: ${\displaystyle \underset{0}{\overset{1}{\int }}\cos \left(\frac{\pi t}{2}\right)}dt$

Applying the substitution, $z=\frac{\pi t}{2}$

Partially differentiate both side,

  $\begin{array}{l}\Rightarrow dz=\frac{\pi }{2}dt\\ \Rightarrow \frac{2}{\pi }dz=dt\end{array}$

Applying limit conversion, putting $t=0$ and $t=1$ in $\frac{\pi t}{2}$ ,

The lower limit converted to $0$ and upper limit converted to $\frac{\pi }{2}$ .

Let’s evaluate the definite integral,

  $\begin{array}{l}={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \left(z\right)\times \frac{2}{\pi }}dz\\ =\frac{2}{\pi }{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\cos \left(z\right)}dz\end{array}$

Therefore, using the common integral,

  $\begin{array}{l}=\frac{2}{\pi }{\left[\sin \left(z\right)\right]}_0^{\frac{\pi }{2}}\\ =\frac{2}{\pi }\left[\sin \left(\frac{\pi }{2}\right)-\sin \left(0\right)\right]\\ =\frac{2}{\pi }\times 1\\ =\frac{2}{\pi }\end{array}$

Hence, The integration of the definite integral ${\displaystyle \underset{0}{\overset{1}{\int }}\cos \left(\frac{\pi t}{2}\right)}dt$ is $\frac{2}{\pi }$ .