Problem 25 - Essential Calculus

Question

Use integration tables to evaluate the integral. $$ \int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+\sin ^{2} x} d x $$

Step-by-Step Solution

Answer

The value of the integral is 0.

Step 1: Substitute variable

Let's start the solution by substituting \(\cos x = t\). This implies that \(-\sin x\, dx = dt\). After this substitution, the integral becomes \[\int \frac{t}{1 + (1 - t^{2})}(-dt)\].

Step 2: Simplify the expression

In this step, simplify the obtained integral. When simplifying, you get \[\int \frac{-t dt}{2 - t^{2}}\].

Step 3: Evaluate the integral

Evaluate the integral using the rule for the integral \(\int \frac{du}{a^{2} - u^{2}} = \frac{1}{a}\ln |\frac{a + u}{a - u}|\) for \(u = t\) and \(a = \sqrt{2}\). After applying this rule, the outcome becomes \[-\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2} + t}{\sqrt{2} - t}\right| + C\].

Step 4: Substitute back the original variable

In this step, replace \(t\) with \(\cos x\). The result is \[-\frac{1}{\sqrt{2}} \ln \left|\frac{\sqrt{2} + \cos x}{\sqrt{2} - \cos x}\right| + C\].

Step 5: Find the definite integral

Finally, find the value for the definite integral. Using the variables of integration we have \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). After applying these limits, the result is 0, because the function is an odd function and the limits are symmetrical around 0.