Problem 20 - Thomas Calculus

Question

Evaluate the integrals. \(\int_{0}^{1 / \sqrt{2}} 2 x \sin ^{-1}\left(x^{2}\right) d x\)

Step-by-Step Solution

Answer

The integral is solved using substitution and integration by parts. The result is derived from applying the limits to the antiderivative.

Step 1: Identify the Substitution

We notice that the problem involves an inverse trigonometric function, which suggests using substitution. Define \( u = x^2 \) and compute \( du \). By differentiating, \( du = 2x \, dx \), which matches the integral's \(2x \, dx\).

Step 2: Change the Limits of Integration

Since \( u = x^2 \), when \( x = 0, \ u = 0^2 = 0 \), and when \(x = \frac{1}{\sqrt{2}}, u = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}\). Thus, the limits of integration for \( u \) change from 0 to \(\frac{1}{2}\).

Step 3: Rewrite the Integral with Substitution

With the substitution \( u = x^2 \) and \( du = 2x \, dx \), the integral becomes \( \int_{0}^{1/2} \sin^{-1}(u) \, du \).

Step 4: Integrate Using Integration by Parts

Utilize the integration by parts formula: \( \int v \, dw = vw - \int w \, dv \). Let \( v = \sin^{-1}(u) \) and \( dw = du \), then \( dv = \frac{1}{\sqrt{1-u^2}} \, du \) and \(w = u\). Therefore, the integral \( \int \sin^{-1}(u) \, du \) becomes \( u\sin^{-1}(u) - \int \frac{u}{\sqrt{1-u^2}} \, du \).

Step 5: Evaluate the Remaining Integral

The remaining integral \( \int \frac{u}{\sqrt{1-u^2}} \, du \) can be solved by substituting \( u = \sin(\theta) \), such that \( du = \cos(\theta) \, d\theta \). The integral becomes \( \int \cos^2(\theta) \, d\theta \), which can be solved using \( \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \). After solving, it yields \( \frac{1}{2}\theta + \frac{1}{4}\sin(2\theta) \). Substituting back gives the antiderivative.

Step 6: Apply Limits to the Antiderivative

After integrating, apply the limits of integration from 0 to \( \frac{1}{2} \). Compute the value of the antiderivative at \( u = \frac{1}{2} \) and subtract the value at \( u = 0 \). This gives the evaluated integral after substitution.

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential when dealing with integrals that involve angles and slopes. These functions, such as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \), are the inverses of the basic trigonometric functions. They help to determine the angle whose trigonometric function value is given. For example:

• If \( y = \sin^{-1}(x) \), then \( x = \sin(y) \).
• The range of \( \sin^{-1}(x) \) is \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

These functions are widely used in calculus, especially in integrating functions that can't be dealt with using standard algebraic techniques. Integration involving inverse trigonometric functions often requires specific techniques like substitution or integration by parts.

Substitution in Integration

Substitution is a powerful technique in integration, used to simplify integrals by changing variables. Fundamentally, when faced with a complex integral, substituting a part of it with a simpler variable can make the integration process manageable. Here's a simple breakdown:

• Identify a part of the integrand (the function you're integrating) that can be substituted with a single variable, e.g., \( u = x^2 \).
• Differentiate the chosen substitution to find \( du \), which is \( 2x \, dx \) in our case, matching another part of the integrand.
• Substitute \( x \), \( dx \), and limits of integration, transforming the integral into a simpler one with respect to \( u \).

Substitution is particularly useful when dealing with inverse trigonometric functions, complex exponential functions, or when the derivative of a potential substitution appears within the integrand. It is a standout tool for addressing integrals that involve nested functions.

Limits of Integration

The limits of integration are the boundaries between which a function is evaluated using definite integrals. Changing variables in an integral often means adapting these limits to the new variable. Here's how this process works:

• Consider the original variable \( x \), with its limits, say from \( 0 \) to \( \frac{1}{\sqrt{2}} \).
• When substituting, as with \( u = x^2 \), calculate the new limits: \( x=0 \) transforms to \( u=0 \), and \( x=\frac{1}{\sqrt{2}} \) changes to \( u=\frac{1}{2} \).
• Use these new limits in the integral rewritten in terms of \( u \).

Properly changing the limits ensures that the definite integral remains consistent with its original context in the \( x \) variable. It's crucial to adjust these when substituting to avoid errors in calculation and obtain the correct area under the curve for the intended variable range.