Problem 43 - Thomas Calculus

Question

Use a substitution to change the integral into one you can find in the table. Then evaluate the integral. \(\int \frac{\sqrt{x}}{\sqrt{1-x}} d x\)

Step-by-Step Solution

Answer

\( \arcsin(\sqrt{x}) - \sqrt{x(1-x)} + C \)

Step 1: Identify the Substitution

To simplify the integral \( \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx \), choose a substitution that will eliminate the square roots. A good choice here is \( x = \sin^2(u) \). This substitution will help manage the square roots using trigonometric identities.

Step 2: Differentiate the Substitution

Using the substitution \( x = \sin^2(u) \), differentiate to find \( dx \):- \( x = \sin^2(u) \) implies \( dx = 2\sin(u)\cos(u) \, du \).- Also note that \( \sqrt{x} = \sin(u) \) and \( \sqrt{1-x} = \cos(u) \) since \( 1 - x = 1 - \sin^2(u) = \cos^2(u) \).

Step 3: Substitute and Simplify the Integral

Express the integral in terms of \( u \):\[ \int \frac{\sqrt{x}}{\sqrt{1-x}} \, dx = \int \frac{\sin(u)}{\cos(u)} \cdot 2\sin(u)\cos(u) \, du \]Simplify the expression:\[ = 2 \int \sin^2(u) \, du \]

Step 4: Use Trigonometric Identity

Replace \( \sin^2(u) \) using the trigonometric identity \( \sin^2(u) = \frac{1 - \cos(2u)}{2} \) to simplify the integral:\[ 2 \int \sin^2(u) \, du = 2 \int \frac{1 - \cos(2u)}{2} \, du = \int (1 - \cos(2u)) \, du \]

Step 5: Integrate

Integrate the simplified expression:\[ \int (1 - \cos(2u)) \, du = \int 1 \, du - \int \cos(2u) \, du \]The integrals are:- \( \int 1 \, du = u \)- \( \int \cos(2u) \, du = \frac{1}{2} \sin(2u) \)So the integrated result is \( u - \frac{1}{2} \sin(2u) + C \), where \( C \) is the constant of integration.

Step 6: Back-Substitute

Replace \( u \) back in terms of \( x \) using \( u = \arcsin(\sqrt{x}) \):- Note that \( \sin(2u) = 2\sin(u)\cos(u) = 2\sqrt{x}\sqrt{1-x} = 2\sqrt{x(1-x)} \).Thus, the solution in terms of \( x \) is:\[ \arcsin(\sqrt{x}) - \sqrt{x(1-x)} + C \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used in integral calculus to simplify integrals involving square roots. The idea is to use trigonometric identities to transform a complex expression into a simpler form. This technique is particularly useful for integrals involving expressions like \( \sqrt{a^2 - x^2} \), \( \sqrt{a^2 + x^2} \), or \( \sqrt{x^2 - a^2} \).

In the given problem, we face an integral with a square root that is not straightforward to integrate directly. By substituting \( x = \sin^2(u) \), the expression \( \sqrt{x} \) becomes \( \sin(u) \) and \( \sqrt{1-x} \) becomes \( \cos(u) \). This transforms the integral into a more manageable form.

The choice of substitution is driven by the identity \( \sin^2(u) + \cos^2(u) = 1 \), which helps to eliminate the square roots. This process reduces the complexity of the integral, paving the way for straightforward application of integration techniques.

Integration Techniques

Integration techniques play a crucial role in solving complex integrals by simplifying them into forms that can be easily integrated. One common technique is trigonometric substitution, as used in our example. By transforming the integral into another form, we often arrive at more standard integrals that are easier to manage.

Once the substitution \( x = \sin^2(u) \) was applied, our integral contained terms \( \sin(u) \) and \( \cos(u) \). We simplified the expression further by recognizing it as \( 2 \int \sin^2(u) \, du \). To integrate \( \sin^2(u) \), we used the identity \( \sin^2(u) = \frac{1 - \cos(2u)}{2} \).

This identity allows the integral to be expressed as \( \int (1 - \cos(2u)) \, du \), a form more amenable to straightforward integration. Using these techniques, we transformed a challenging integral into something approachable using basic antiderivatives.

Definite Integrals

While the current problem addresses an indefinite integral, it's useful to connect it to definite integrals to understand broader applications. Definite integrals evaluate the accumulation of quantities across specific intervals, giving a total value or net change.

For example, if our transformed integral was calculated within set limits or bounds, we would be computing a definite integral. The process involves substituting these bounds into the antiderivative and finding the difference between the upper and lower evaluation.

When dealing with trigonometric substitution in definite integrals, the substitution process changes the bounds of integration. After finding the antiderivative, return to the original variable by reversing the substitution, applying the new integral limits. This careful attention to the substitution and its limits ensures accuracy in finding the area or value needed from the definite integral.