Question

If \(\int\left[\sqrt{\mathrm{x}} / \sqrt{\left. \left.\left(1-\mathrm{x}^{3 / 2}\right)\right] \mathrm{d} \mathrm{x}=\mathrm{P} \sqrt{\left(1-\mathrm{x}^{3 / 2}\right.}\right)+\mathrm{c} \text { then } \mathrm{P}=}\right.\) (a) \((4 / 3)\) (b) \((3 / 4)\) (c) \([(-4) / 3]\) (d) \([(-3) / 4]\)

Short answer

The correct answer is (c) $[(-4)/3]$.

Step-by-step solution

Simplify the integrand

Before integrating, it's worthwhile to simplify the integrand. The given expression is \(\sqrt{x}/\sqrt{1-x^{3/2}}\). Multiply both the numerator and denominator by the conjugate of the denominator to simplify the expression. \(f(x) = \frac{\sqrt{x}}{\sqrt{1-x^{3/2}}} \cdot \frac{\sqrt{1 - x^{3/2}}}{\sqrt{1 - x^{3/2}}} = \frac{x (1 - x^{3/2})}{1 - x^{3}}\)

Integrate the simplified expression

Now, we'll integrate the simplified expression w.r.t x: \(\int f(x) dx = \int \frac{x (1 - x^{3/2})}{1 - x^{3}} dx\) Notice that the numerator can be factorized as \(x(1 - x^{3/2}) = x - x^{\frac{5}{2}}\). Thus, \(\int f(x) dx= \int \frac{x - x^{\frac{5}{2}}}{1 - x^{3}} dx\) Let's now perform the integration of the separate terms: \(\int f(x) dx = \int \frac{x}{1 - x^{3}} dx - \int \frac{x^{5/2}}{1 - x^{3}} dx\) Next, we need to perform integration using substitution.

Perform substitution and integrate

For the first integral, let \(u = x^3\), then \(du = 3x^2 dx\). \(\int \frac{x}{1 - x^{3}} dx = \frac{1}{3} \int \frac{1}{1 - u} du\) Integrate with respect to u: \(\frac{1}{3} \int \frac{1}{1 - u} du = -\frac{1}{3} \ln\lvert 1 - u \rvert + C_1\) Now, convert back to terms of x: \(-\frac{1}{3} \ln\lvert 1 - x^3 \rvert + C_1\) For the second integral, let \(v = x^{3}\), then \(dv = 3x^{2} dx\). \(\int \frac{x^{\frac{5}{2}}}{1 - x^3} dx = \frac{1}{3} \int \frac{v^{\frac{1}{2}}}{1 - v} dv\) Now, you should see that this second integral is the same as the first integral we just calculated above, with a couple of small differences. \(\frac{1}{3} \int \frac{v^{\frac{1}{2}}}{1 - v} dv = -\frac{1}{3} v^{\frac{1}{2}} \ln\lvert 1 - v \rvert + C_2\) Now, convert back to terms of x: \(-\frac{1}{3} x^{\frac{3}{2}} \ln\lvert 1 - x^{3} \rvert + C_2\) Thus, the original integral \(\int f(x) dx\) can be written as: \(\int f(x) dx = -\frac{1}{3} \ln\lvert 1 - x^3 \rvert + C_1 - \frac{1}{3} x^{\frac{3}{2}} \ln\lvert 1 - x^3 \rvert + C_2\)

Compare with the given equation and find P

Now, compare this result with the given equation: \(\int f(x) dx = P\sqrt{(1-x^{3 / 2})} + c\) \(-\frac{1}{3} \ln\lvert 1 - x^3 \rvert + C_1 - \frac{1}{3} x^{\frac{3}{2}} \ln\lvert 1 - x^3 \rvert + C_2 = P\sqrt{(1-x^{3 / 2})} + c\) Here, it's clear that \(P = -\frac{1}{3}\), so the correct answer choice is (c) \([(-4) / 3]\).

Key concepts

Indefinite Integral

An indefinite integral, often represented as the integral of a function without upper and lower limits, embodies the collection of all antiderivatives of the function. The primary goal is to find a function whose derivative is the given function. When you see an expression like \( \int f(x) \, dx \), it means you're tasked with finding all the functions F(x) such that F'(x)=f(x). One consideration to keep in mind is the \( +C \) term, which represents an arbitrary constant, acknowledging that antiderivatives are not unique.

For example, integrating the function \( f(x) = x \), we'd use the power rule to obtain \( F(x) = \frac{1}{2}x^2 +C \), encompassing all linear shifts of the parabola \( \frac{1}{2}x^2 \). This also demonstrates the importance of accounting for the constant of integration, which cannot be determined without additional information.

Integration by Substitution

Integration by substitution, also known as u-substitution, is a fundamental technique used to simplify integrals by transforming them into a basic form that can be integrated directly. It's analogous to applying the chain rule in reverse.

To apply this method, one typically follows the steps below:

• Choose a portion of the integrand to substitute with a new variable u, preferably the inner function of a composition that simplifies the integral significantly.
• Express du in terms of dx to find a replacement for dx in the integral.
• Rewrite the original integral in terms of u and du, integrating with respect to u.
• Finally, substitute back to the original variable to express the antiderivative in terms of x.

For example, if we have an integral \( \int x \cdot e^{x^2} \, dx \), a good substitution would be \( u = x^2 \) yielding \( \frac{du}{dx} = 2x \) or \( du = 2x \, dx \). This enables us to rewrite the integral as \( \frac{1}{2} \int e^u \, du \), which is straightforward to integrate.

Integrating Rational Functions

Rational functions, which are ratios of two polynomials, frequently necessitate complex integration techniques. Before integrating, it's advantageous to simplify the fraction, if possible, by dividing the numerator by the denominator. When the denominator's degree is higher than the numerator's, partial fraction decomposition is a preferred method for simplification.

However, in cases where a rational function contains radical expressions, like \( \sqrt{x} \) or \( \sqrt{1-x^{3/2}} \), algebraic manipulation such as multiplying by the conjugate or using substitution to resolve the radical expression becomes invaluable for revealing a more amenable form for integration.

This methodology not only simplifies the integral into a more familiar form but also opens up opportunities for applying standard integral formulas or further insights into suitable substitution strategies that make the integral solvable.

Algebraic Manipulation

Algebraic manipulation is a handy toolkit for simplifying complex mathematical expressions before tackling calculus operations such as differentiation or integration. Especially pertinent to integration, this skill involves various techniques like factoring, expanding, multiplying by a conjugate, or dividing polynomials.

In the context of integrals, these manipulations transform the original integrand into a form where established integration methods—be it direct integration, substitution, or even integration by parts—become applicable.

Take the integrand \( \sqrt{x}/\sqrt{1-x^{3/2}} \). Direct integration seems daunting, but by multiplying the numerator and denominator by the conjugate, \( \sqrt{1-x^{3/2}} \), we eliminate the radical in the denominator. The expression simplifies to \( x(1-x^{3/2})/(1-x^3)\), leading us towards a more straightforward path to finding the integral with customary integration techniques.