Problem 19 - Essential Calculus

Question

Evaluate the integral. $$ \int_{0}^{1 / \sqrt{2}} \frac{\arcsin x}{\sqrt{1-x^{2}}} d x $$

Step-by-Step Solution

Answer

\(\frac{\pi^2}{32}\)

Step 1: Substitution

First, we do the substitution \(\ x = \sin \theta \), and therefore \( dx = \cos \theta d\theta \). Also, by substituting the limits, we get new limits of integration: for x=0, \(\theta=0\), and for \(x=1/\sqrt{2}\), \(\theta=\pi/4\).

Step 2: Substitute into the integral

Substitute \(x\), \( dx\) and the limits into the integral: \(\int_{0}^{\pi/4} \frac{\arcsin{(\sin{\theta})}}{\sqrt{1-\sin^2{\theta}}} \cos{\theta}\, d\theta\). Here, \(\arcsin{(\sin{\theta})}\) simplifies to \(\theta\) and \(\sqrt{1-\sin^2{\theta}}\) simplifies to \(\cos{\theta}\).

Step 3: Perform the integration

Now the integral simplifies to \(\int_{0}^{\pi/4} \theta \, d\theta\). This is now a straightforward integral. The antiderivative of \(\theta\) is \( \frac{1}{2} \theta^2\).

Step 4: Apply the limits

Using the properties of definite integrals, we substitute the limits into the antiderivative: \(\frac{1}{2} (\pi/4)^2 - \frac{1}{2} (0)^2\).

Step 5: Simplify the final expression

Cleaning up the expression, we get \( \frac{\pi^2}{32}\). So, the value of the integral \(\int_{0}^{1 / \sqrt{2}} \frac{\arcsin x}{\sqrt{1-x^{2}}} d x\) is \( \frac{\pi^2}{32}\).