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Chapter 12: Problem 1066

The value of the integral \((1 / 2) \int_{0}\left[d x /\left\\{(1-x)^{(3 / 2)} \sqrt{(1+x)\\}]}\right.\right.\) is.... (a) 0 (b) \((1 / 2)\) (c) \(\sqrt{3}-1\) (d) 2

Short Answer

Expert verified
None of the given options matches our result, which indicates that there might be an issue with the problem statement or the options provided.

Step by step solution

01

Substitution

Let's try a substitution to simplify the integral. Choose: \(u = 1 - x\) Then, differentiating with respect to x, we get: \(du = -dx\) Now, substitute u back into the integral and change the limits of integration according to the substitution: \[\frac{1}{2} \int_{0}^{1}\frac{d x}{(1-x)^{(3 / 2)}\sqrt{(1+x)}} = -\frac{1}{2} \int_{1}^{0}\frac{du}{(u)^{(3 / 2)}\sqrt{(2-u)}}\]
02

Change the limits of integration

Now, reverse the limits of integration and get rid of the negative sign: \[\frac{1}{2} \int_{1}^{0}\frac{du}{(u)^{(3 / 2)}\sqrt{(2-u)}} = \frac{1}{2} \int_{0}^{1}\frac{du}{u^{\frac{3}{2}}\sqrt{2-u}}\]
03

Use substitution again

Use another substitution to simplify the integral further: \(v = 2 - u\) Then, differentiating with respect to u, we have: \(dv = -du\) Now substitute v back into the integral and change the limits of integration as per the substitution: \[\frac{1}{2}\int_{0}^{1}\frac{du}{u^{\frac{3}{2}}\sqrt{2-u}} = -\frac{1}{2} \int_{2}^{1}\frac{dv}{(2-v)^{\frac{3}{2}}\sqrt{v}}\]
04

Evaluate the integral

Recall that the integral of \(\frac{1}{\sqrt{a^2 - x^2}}\) is \(\arcsin\frac{x}{a} + C\). Our integral resembles this form where a = 2. Reverse the limits of integration and get rid of the negative sign: \[-\frac{1}{2} \int_{2}^{1}\frac{dv}{(2-v)^{\frac{3}{2}}\sqrt{v}} = \frac{1}{2} \int_{1}^{2}\frac{dv}{(2-v)^{\frac{3}{2}}\sqrt{v}}\] Now, evaluate the integral: \[\frac{1}{2} \int_{1}^{2}\frac{dv}{(2-v)^{\frac{3}{2}}\sqrt{v}} = \frac{1}{2} \left[\arcsin\left(\frac{v}{2}\right)\right]_{1}^{2}\]
05

Find the value of the definite integral

Place the limits of integration into the expression: \[\frac{1}{2} \left[\arcsin\left(\frac{v}{2}\right)\right]_{1}^{2} = \frac{1}{2}\left(\arcsin\left(\frac{2}{2}\right) - \arcsin\left(\frac{1}{2}\right)\right)\] \[\frac{1}{2}(\arcsin(1) - \arcsin\left(\frac{1}{2}\right)) = \frac{1}{2}\left(\frac{\pi}{2} - \frac{\pi}{6}\right)\]
06

Simplify and match with the given options

Simplify the expression: \[\frac{1}{2}\left(\frac{\pi}{2} - \frac{\pi}{6}\right) = \frac{1}{2}\left(\frac{3\pi - \pi}{6}\right) = \frac{1}{2}\left(\frac{2\pi}{6}\right) = \frac{\pi}{6}\] Now compare this result with the given options: (a) 0 (b) \(\frac{1}{2}\) (c) \(\sqrt{3}-1\) (d) 2 None of the given options matches our result, which indicates that there might be an issue with the problem statement or the options provided.

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