Problem 42 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^{1/\sqrt{2}}_{1/2} \frac{4}{\sqrt{1 - x^2}} \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{\pi}{3} \).

Step 1: Recognize the Integral as an Inverse Trigonometric Function

The integral given is \( \int \frac{4}{\sqrt{1-x^2}}\, dx \). The expression inside the integral matches the standard form for the derivative of the arcsin function, \( \frac{1}{\sqrt{1-x^2}} \). Therefore, we can rewrite the integral as \( 4 \cdot \arcsin(x) \).

Step 2: Set up the Definite Integral

We need to evaluate the integral \( \int^{1/\sqrt{2}}_{1/2} 4 \cdot \arcsin(x)\, dx \). Applying the Fundamental Theorem of Calculus, the definite integral becomes:\[ 4 \cdot [ \arcsin(x) ]_{1/2}^{1/\sqrt{2}}. \]

Step 3: Evaluate the Arc Sine at the Boundaries

First, evaluate \( \arcsin(x) \) at the upper limit, \( x = \frac{1}{\sqrt{2}} \). We know from trigonometric identities that \( \arcsin\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \). Now, evaluate \( \arcsin(x) \) at the lower limit, \( x = \frac{1}{2} \), which gives \( \arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6} \).

Step 4: Compute the Definite Integral

Substitute these results into the evaluated integral:\[ 4 \cdot \left[ \frac{\pi}{4} - \frac{\pi}{6} \right]. \]

Step 5: Simplify the Result

Calculate the difference inside the brackets: \( \frac{\pi}{4} - \frac{\pi}{6} = \frac{3\pi}{12} - \frac{2\pi}{12} = \frac{\pi}{12} \).Now multiply by 4 to get:\[ 4 \cdot \frac{\pi}{12} = \frac{\pi}{3}. \]

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions help us reverse the process of a trigonometric function. Imagine you know the sine, cosine, or tangent of an angle, and you want to find the angle itself. That's where these functions come into play! Some of the common inverse trigonometric functions are arcsin, arccos, and arctan.

These inverse functions are essential in calculus for integrating certain types of expressions. For instance, the integral \( \int \frac{1}{\sqrt{1-x^2}} \, dx \) is recognized as the antiderivative of \( \arcsin(x) \). This is because the derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \). Notice the pattern: the structure inside the integral matches what's inside the square root. So by "going backward," we identify the form of \( \arcsin(x) \) in our solution. This knowledge is crucial when approaching problems requiring integration of trigonometric forms. Recognizing these patterns will save you time and confusion when executing integral calculus exercises.

Definite Integrals

Definite integrals calculate the signed area under a curve, between two specified limits. In the given problem, the definite integral \( \int^{1/\sqrt{2}}_{1/2} 4 \cdot \arcsin(x)\, dx \) specifies the interval over which we evaluate the area. The outcome is a single number representing this area.

To handle the definite integral, we set it up as a difference of the antiderivative evaluated at the upper and lower limits. This involves a specific process:

• First, find the antiderivative function (the integrand here was identified as \( 4 \cdot \arcsin(x) \)).
• Then, plug the limits into this antiderivative.
• Subtract the value obtained at the lower limit from the value obtained at the upper limit.

For our example, this results in \( \arcsin \left( \frac{1}{\sqrt{2}} \right) - \arcsin \left( \frac{1}{2} \right) \). Don't forget the coefficient; it's crucial for the final calculation!

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus links the concept of differentiation with that of integration. It has two main parts, but for definite integrals like our example, we focus on the second part. This part states that if you can find an antiderivative \( F(x) \) for your function \( f(x) \), you can compute the definite integral \( \int_{a}^{b} f(x) \, dx \) by evaluating \( F(b) - F(a) \). This link helps you directly evaluate the integral, turning a problem of finding the area under a curve into mere subtraction.

In our problem, after finding the antiderivative \( 4 \cdot \arcsin(x) \), we simply evaluated this expression at the upper and lower limits. Applying this theorem simplifies the process of integration and provides a straightforward route to the solution. Without the Fundamental Theorem of Calculus, computations involving integral calculus would be much more complicated and time-consuming.