Question

In Exercises \(53-66,\) make a \(u\) -substitution and integrate from \(u(a)\) to \(u(b) .\) $$\int_{-\pi}^{\pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x$$

Short answer

The result of the integral \(\int_{-\pi}^{\pi} \frac{\cos x}{\sqrt{4+3 \sin x}} d x\) is 0.

Step-by-step solution

Recognition of 'u'

To solve this integral, recognize a function that in its derivative form is present in the integrand. In this case, let's make the substitution \(u = \sin{x}\).

Compute the derivative of 'u'

Compute the derivative of the substitution \(u = \sin{x}\). The derivative \(du = \cos{x}\; dx\). This derivative simplifies the original integral.

Convert the integral from x to u

Substitute values into the integral. In this case, change the integral of \(x\) to that of \(u\) by substitifying \(\cos{x}\; dx\) for \(du\). This leads to \(\int_{-\pi}^{\pi} \frac{\cos x}{\sqrt{4+3 u}} dx = \int_{u(-\pi)}^{u(\pi)} \frac{1}{\sqrt{4+3u}} du\).

Compute the limits in terms of 'u'

Compute the limits as 'u' of the given 'x' values \(-\pi\) and \(\pi\). We have \(u(-\pi) = \sin{(-\pi)} = 0\) and \(u(\pi) = \sin{\pi} = 0\).

Solve the integral

The integral is now in the form \(\int_{0}^{0} \frac{1}{\sqrt{4+3u}} du\). Integrating this expression using standard techniques (such as recognizing this as a standard inverse trigonometric integral), we get the result as 0, since the limits of the integral are the same.