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Chapter 4: Problem 20

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

Short Answer

Expert verified
The evaluation of the given integral is \(-(1/32 \pi^2 - 1/8 \pi^2) = -3/32 \pi^2\)

Step by step solution

01

Variable substitution

Perform the substitution \(u = \arccos x\), hence \(du = -\frac{1}{\sqrt{1-x^2}} dx\). Flip both sides to get \(dx = - \sqrt{1-x^2} du\). Also change the integral limits as per the new variable \(u\). When \(x = 0\), \(u = \arccos 0 = \frac{\pi}{2}\), and when \(x = 1/\sqrt{2}\), \(u = \arccos 1/\sqrt{2} = \frac{\pi}{4}\). This will change our integral into the following form: \(- \int_{\pi /2}^{\pi /4} u du\) .
02

Evaluate the integral

This is a simple integral now. The integral of \(u\) with respect to \(u\) is \(1/2 u^2\). Now we need to evaluate this from \(\pi/4\) to \(\pi/2\).
03

Apply limits and simplify

The final step is to plug in our limits of integration and subtract: \(-[1/2 {(\pi/4)}^2 - 1/2 {(\pi/2)}^2 ]\). Simplify this expression to get final answer

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