Problem 38 - Calculus: One and Several Variables

Question

Calculate. $$\int_{0}^{1 / 2} \frac{1+x}{\sqrt{1-x^{2}}} d x$$

Step-by-Step Solution

Answer

The short answer for the definite integral is: \(\int_{0}^{1 / 2} \frac{1+x}{\sqrt{1-x^{2}}} d x = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1\).

Step 1: Identify a suitable substitution

The function \(f(x) = \frac{1+x}{\sqrt{1-x^2}}\) contains the term under the square root \(\sqrt{1-x^2}\). This suggests we should use a trigonometric substitution to rewrite the expression, as these substitutions are designed to handle expressions that involve the square root of a sum or difference of squares. Specifically, we can use the substitution \(x = \sin{u}\).

Step 2: Perform the substitution

Given that \(x = \sin{u}\), we can differentiate both sides with respect to \(u\) to obtain \(dx = \cos{u} \ du\). Now, we can make the following substitutions: - Replace \(x\) with \(\sin{u}\) - Replace \(\sqrt{1-x^2}\) with \(\cos{u}\) - Replace \(dx\) with \(\cos{u} \ du\) Doing so, we get the following expression for our integral: $$\int_{0}^{1 / 2} \frac{1+x}{\sqrt{1-x^{2}}} d x = \int \frac{1+\sin{u}}{\cos{u}}\cdot \cos{u} du$$ Now, we need new limits of integration. When \(x = 0\), we have \(u = \sin^{-1}(0) = 0\). When \(x = \frac{1}{2}\), we have \(u = \sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}\). Therefore, our new limits of integration are \(0\) and \(\frac{\pi}{6}\). The integral now becomes: $$\int_{0}^{\frac{\pi}{6}} \frac{1+\sin{u}}{\cos{u}}\cdot \cos{u} du$$

Step 3: Simplify the integral

Now, we can simplify the integral by canceling the \(\cos{u}\) terms: $$\int_{0}^{\frac{\pi}{6}} (1+\sin{u}) du$$

Step 4: Calculate the antiderivative

To evaluate the integral, we find the antiderivative: $$\int (1+\sin{u}) du = u - \cos{u} + C$$

Step 5: Use the Fundamental Theorem of Calculus

Now we can use the Fundamental Theorem of Calculus to evaluate the definite integral with the updated limits of integration: $$\int_{0}^{\frac{\pi}{6}} (1+\sin{u}) du = \left[u - \cos{u}\right]_{0}^{\frac{\pi}{6}} = \left[\left(\frac{\pi}{6} - \cos{\frac{\pi}{6}}\right) - \left(0 - \cos{0}\right)\right]$$

Step 6: Evaluate and simplify

We can now evaluate and simplify the expression: $$\left(\frac{\pi}{6} - \cos{\frac{\pi}{6}}\right) - \left(0 - \cos{0}\right) = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1$$ Now, we have the final answer: $$\int_{0}^{1 / 2} \frac{1+x}{\sqrt{1-x^{2}}} d x = \frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1$$

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique often used in integral calculus to simplify integrals involving square roots of quadratic expressions. This method leverages the identities and relationships in trigonometric functions to transform complex algebraic expressions into simpler trigonometric forms. For the function \(f(x) = \frac{1+x}{\sqrt{1-x^2}}\), we notice the term \(\sqrt{1-x^2}\), which fits the identity \(\sin^2{u} + \cos^2{u} = 1\). Therefore, by substituting \(x = \sin{u}\), we convert the integral into a more manageable form.
The steps for solving using trigonometric substitution include:

• Select a trigonometric function as the substitute variable.
• Replace \(x\), \(\sqrt{1-x^2}\), and \(dx\) accordingly.
• Adjust the limits of integration to reflect this substitution.

This substitution reduces the complexity by using known derivatives and antiderivatives of trigonometric functions, making it easier to evaluate.

Definite Integrals

Definite integrals are used to calculate the total accumulation of a quantity, expressed by an integral, within a specified interval. In this exercise, we evaluate the integral \(\int_{0}^{1 / 2} \frac{1+x}{\sqrt{1-x^{2}}} dx\) by transforming it through substitution, and then integrating over the specific limits.
Key aspects to keep in mind with definite integrals:

• The limits of integration must be adjusted to new variables when substitutions are made.
• The result is a number representing the net area under the curve over the indicated interval.
• Upon finding the antiderivative, apply the limits to find the exact value of the definite integral.

This method provides not only the total change but, in certain cases such as area, an actual geometric interpretation or meaning for the evaluated integral.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus forms a bridge between differential and integral calculus. It provides a method to evaluate a definite integral by using the antiderivative of the function. The theorem states that if \(F(x)\) is an antiderivative of \(f(x)\), then the definite integral from \(a\) to \(b\) of \(f(x)\) is given by \(F(b) - F(a)\).
When applying this theorem:

• Find an antiderivative \(F(u)\).
• Substitute the upper and lower limits into \(F(u)\).
• Subtract the lower limit evaluation from the upper limit evaluation.

In the exercise, the antiderivative \(F(u) = u - \cos{u} + C\) was found for \(\int (1+\sin{u}) du\). Using the limits \(0\) to \(\frac{\pi}{6}\), applying the Fundamental Theorem of Calculus provided a final result of \(\frac{\pi}{6} - \frac{\sqrt{3}}{2} + 1\). This showcases how powerful and essential this theorem is in evaluating integrals.