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Chapter 12: Problem 1040

The value of integral \({ }^{1} \int_{0}\left[1 /\left\\{1-\mathrm{x}+\sqrt{\left. \left.\left(2 \mathrm{x}-\mathrm{x}^{2}\right)\right\\}\right] \mathrm{d} \mathrm{x} \text { is...... }}\right.\right.\) (a) 1 (b) \((1 / 2)\) (c) \((\pi / 4)\) (d) \((\pi / 2)\)

Short Answer

Expert verified
The short answer to the integral is: \[\int_0^1 \frac{1}{\sqrt{2x - x^2}} dx = \frac{\pi}{2}\]. Therefore, the correct answer is (d).

Step by step solution

01

Identify an Integration Technique

Since the given integral contains a square root in the denominator, we will use substitution technique. The goal is to get rid of the square root and rewrite the integral in a more manageable form.
02

Perform the Substitution

Let's use the substitution \(x = 1 - \sin{u}\), which means \(dx = -\cos{u} du\). Now, let's rewrite the integral in terms of "\(u\)": \(\) \[ Expressing x in terms of u, we have - {x^2} + 2x = - (1 - \sin{u})^2 + 2(1-\sin{u}) \] Now due to sin^2 + cos^2 = 1 we get (1 - {sin^2(u)}) and now replacing the function in the integral we get, \( -\int \frac{1}{\sqrt{1 - \sin^2{u}}} \cos{u} du \)
03

Apply Trigonometric Identity

Now we apply the trigonometric identity: \( \sin^2{u} + \cos^2{u} =1 \) by rewriting \(\sqrt{1 - \sin^2{u}}\) in terms of cosine: \( = -\int \frac{1}{\cos{u}} \cos{u} du \)
04

Simplify the Integral

Now we can simplify the integral expression: \( -\int 1 du \)
05

Evaluate the Integral

Now we can evaluate the integral: \(-\int 1 du = -u + C\)
06

Reverse the Substitution

Now let's reverse the substitution \(x = 1 - \sin{u}\) and replace "\(-u + C\)" with the expression in terms of x: \( = -( \arcsin{(1-x)}) + C \)
07

Evaluate the Definite Integral

Now evaluate the definite integral by plugging in the limits x = 0 to x = 1: \(-( \arcsin{(1-1)}) - (\arcsin{(1-0)}) = -(\arcsin{0} - \arcsin{1})\) \(-(0-\frac{\pi}{2}) = \frac{\pi}{2}\) Thus, the integral evaluates to \( \frac{\pi}{2}\), which corresponds to the answer choice (d).

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