Chapter 7.2, Problem 38E

To determine

To evaluate the integral .

Answer to Problem 38E

The value of the ${\displaystyle \int \frac{6\cos t}{{\left(2+\sin t\right)}^2}}dt=-\frac{6}{2+\sin t}+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \frac{6\cos t}{{\left(2+\sin t\right)}^2}}dt$

Calculation:

The integral is ${\displaystyle \int \frac{6\cos t}{{\left(2+\sin t\right)}^2}}dt$

Using substitution method of integration

  $\begin{array}{l}{\displaystyle \int \frac{6\cos t}{{\left(2+\sin t\right)}^2}}dt\\ \text{let }2+\sin t=u\\ \text{differentiating both side}\\ \cos tdt=du\\ dt=\frac{du}{\cos t}\\ {\displaystyle \int \frac{6\cos t}{{\left(2+\sin t\right)}^2}}dt\\ {\displaystyle \int \frac{6\cos t}{{\left(u\right)}^2}}\times \frac{du}{\cos t}\\ 6{\displaystyle \int \frac{1}{u^2}}du=-\frac{6}{u}+C=-\frac{6}{2+\sin t}+C,\text{where }C\text{ is constant}\text{.}\\ \end{array}$

Hence ,the value of the ${\displaystyle \int \frac{6\cos t}{{\left(2+\sin t\right)}^2}}dt=-\frac{6}{2+\sin t}+C$