Chapter 17: Problem 6

By means of the substitution \(x=\sin ^{2} \theta\) find \(\int \sqrt{\frac{x}{1-x}} \mathrm{~d} x\).

Short Answer

Expert verified

The result of the integration is \(\frac{1}{2}\left(\arcsin\sqrt{x} - \sqrt{x}\sqrt{1-x}\right) + C\).

Step by step solution

Find the differential \(\mathrm{d}x\) by differentiating the substitution equation with respect to \(\theta\)

Differentiate the substitution equation \(x = \sin^2\theta\) with respect to \(\theta\) to find \(\mathrm{d}x\): \(\frac{d}{d\theta}(x) = \frac{d}{d\theta}(\sin^2\theta)\) Therefore, \(\mathrm{d}x = 2\sin\theta\cos\theta \, \mathrm{d}\theta\)

Replace \(x\) and \(\mathrm{d}x\) in the integral using the substitution equation and the result from Step 1

Substitute \(x\) with \(\sin^2\theta\) and \(\mathrm{d}x\) with \(2\sin\theta\cos\theta \, \mathrm{d}\theta\) in the integral to get: \(\int \sqrt{\frac{x}{1-x}} \mathrm{~d} x = \int \sqrt{\frac{\sin^2\theta}{1-\sin^2\theta}}(2\sin\theta\cos\theta) \, \mathrm{d}\theta\)

Simplify the integral using trigonometric identities

Notice that the integral can be simplified using the trigonometric identity \(1 - \sin^2\theta = \cos^2\theta\): \(\int \sqrt{\frac{\sin^2\theta}{1-\sin^2\theta}}(2\sin\theta\cos\theta) \, \mathrm{d}\theta = \int \sqrt{\frac{\sin^2\theta}{\cos^2\theta}}(2\sin\theta\cos\theta) \, \mathrm{d}\theta\) Now, rewrite the integral using properties of square roots: \(\int \sqrt{\frac{\sin^2\theta}{\cos^2\theta}}(2\sin\theta\cos\theta) \, \mathrm{d}\theta = \int \frac{\sin\theta}{\cos\theta}(2\sin\theta\cos\theta) \, \mathrm{d}\theta\) _

Evaluate the integral with the given substitution

Integrate with respect to \(\theta\) and simplify: \(\int \frac{\sin\theta}{\cos\theta}(2\sin\theta\cos\theta) \, \mathrm{d}\theta = 2\int \sin^2\theta \, \mathrm{d}\theta\) Now, use another trigonometric identity, the double-angle formula for cosine: \(\cos{2\theta} = 1 - 2\sin^2\theta\), to rewrite the integral: \(\int \sin^2\theta \, \mathrm{d}\theta = \frac{1}{2}\int (1 - \cos{2\theta}) \, \mathrm{d}\theta\) Integrate the expression: \(\frac{1}{2}\int (1 - \cos{2\theta}) \, \mathrm{d}\theta = \frac{1}{2}(\theta - \frac{1}{2}\sin{2\theta}) + C\)

Convert back to the \(x\) variable

Replace \(\theta\) and \(\sin{2\theta}\) with the corresponding expression in terms of the \(x\) variable using the substitution equation \(x = \sin^2\theta\) and the double-angle formula for sine \(\sin{2\theta} = 2\sin\theta\cos\theta\): \(\frac{1}{2}(\theta - \frac{1}{2}\sin{2\theta}) + C = \frac{1}{2}\left(\arcsin\sqrt{x} - \frac{1}{2}(2\sqrt{x}\sqrt{1-x})\right) + C = \frac{1}{2}\left(\arcsin\sqrt{x} - \sqrt{x}\sqrt{1-x}\right) + C\) So, the final answer is: \(\int \sqrt{\frac{x}{1-x}} \mathrm{~d} x = \frac{1}{2}\left(\arcsin\sqrt{x} - \sqrt{x}\sqrt{1-x}\right) + C\)

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By means of the substitution \(x=\sin ^{2} \theta\) find \(\int \sqrt{\frac{x}{1-x}} \mathrm{~d} x\).

Short Answer

Step by step solution

Find the differential \(\mathrm{d}x\) by differentiating the substitution equation with respect to \(\theta\)

Replace \(x\) and \(\mathrm{d}x\) in the integral using the substitution equation and the result from Step 1

Simplify the integral using trigonometric identities

Evaluate the integral with the given substitution

Convert back to the \(x\) variable

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