Chapter 7.2, Problem 31E

To determine

To evaluate the integral .

Answer to Problem 31E

The value of the ${\displaystyle \int \cos \left(3z+4\right)dz}=\frac{1}{3}\sin \left(3z+4\right)+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \cos \left(3z+4\right)dz}$

Calculation:

The integral is ${\displaystyle \int \cos \left(3z+4\right)dz}$

Using substitution method of integration

  $\begin{array}{l}{\displaystyle \int \cos \left(3z+4\right)dz}\\ \text{let }3z+4=t\\ \text{differentiating both side}\\ 3dz=dt\\ dz=\frac{dt}{3}\\ {\displaystyle \int \frac{\cos t}{3}}dt\\ \frac{1}{3}{\displaystyle \int \cos tdt}=\frac{1}{3}\sin t+C=\frac{1}{3}\sin \left(3z+4\right)+C,\text{where C is constant}\text{.}\\ \end{array}$

Hence ,the value of the ${\displaystyle \int \cos \left(3z+4\right)dz}=\frac{1}{3}\sin \left(3z+4\right)+C$