Problem 22 - Essential Calculus

Question

Evaluate the integral. $$ \int_{0}^{\pi / 2} \frac{\cos x}{1+\sin ^{2} x} d x $$

Step-by-Step Solution

Answer

The value of the integral is \(\pi / 4\).

Step 1: Setting Up the Substitution

In attempt to simplify the integral, let's make a substitution. We substitute \(u = \sin x\) which implies \(du = \cos x \, dx\). Replacing \(\cos x \, dx\) with \(du\) makes the integral simpler. But we need to make sure to change the limits of integration as well. The initial lower limit was \(x = 0\) which now translates to \(u = \sin 0 = 0\), and the initial upper limit was \(x = \pi / 2\) which now translates to \(u = \sin (\pi / 2) = 1\).

Step 2: Substitute Into the Integral

Now our integral, with the substitution and the new limits of integration, becomes: \(\int_{0}^{1} \frac{1}{1+u^2} du\). This is a simpler integral to calculate.

Step 3: Evaluating the Integral

The integral \(\int \frac{1}{1+u^2} du\) is a standard integral whose result is \(\arctan(u)\). Applying this rule, we get \(\arctan(u)\) from 0 to 1.

Step 4: Evaluating the Limits

The final step involves substituting the limits of integration back into the result, and subtracting to find the final answer: \(\arctan(1) - \arctan(0) = \pi/4 - 0 = \pi/4\).