Chapter 5.4, Problem 8E

To determine

To find: The value of $dy/dx$ .

Answer to Problem 8E

The answer is: $\frac{dy}{dx}=\frac{2-\sin x}{3+\cos x}$

Explanation of Solution

Given information:

The expression is: $y={{\displaystyle \int }}_{-\pi }^{\pi }\frac{2-\sin t}{3+\cos t}dt$

Calculation:

The given expression is: $y={{\displaystyle \int }}_{-\pi }^{\pi }\frac{2-\sin t}{3+\cos t}dt$

The calculus fundamental theorem, part 1:

If $f$ is continuous on $[a,b]$ and

  $F(x)={{\displaystyle \int }}_a^{x}f(t)dt$

The derivative of $F(x)$ is:

  $\begin{array}{c}{F}^{\prime }(x)=\frac{d}{dx}\left[{{\displaystyle \int }}_a^{x}f(t)dt\right]\\ =f(x)\end{array}$

It follows that,

  $y={{\displaystyle \int }}_{-\pi }^{\pi }\frac{2-\sin t}{3+\cos t}dt$

Because discriminate with respect to the upper limit in this situation and the lower limit is a constant, write

  $\frac{dy}{dx}=\frac{2-\sin x}{3+\cos x}$

Therefore the required value of the given expression $y={{\displaystyle \int }}_{-\pi }^{\pi }\frac{2-\sin t}{3+\cos t}dt$ is

  $\frac{dy}{dx}=\frac{2-\sin x}{3+\cos x}$ .