Problem 48 - Calculus Single Variable

Question

Calculate the given integral by first integrating by parts and then making a trigonometric substitution. $$ \int_{0}^{1 / \sqrt{2}} x \arcsin (x) d x $$

Step-by-Step Solution

Answer

The integral evaluates to 0.

Step 1: Choose Parts for Integration by Parts

To apply integration by parts, identify functions to be differentiated and integrated. Let \( u = \arcsin(x) \) and \( dv = x \, dx \). Therefore, \( du = \frac{1}{\sqrt{1-x^2}} \cdot dx \) and \( v = \frac{x^2}{2} \).

Step 2: Apply Integration by Parts Formula

Integration by parts is given by \( \int u \, dv = uv - \int v \, du \). Substitute the terms: \[ \int_{0}^{1/\sqrt{2}} x \arcsin(x) \, dx = \left[ \frac{x^2}{2} \arcsin(x) \right]_0^{1/\sqrt{2}} - \int_{0}^{1/\sqrt{2}} \frac{x^2}{2\sqrt{1-x^2}} \, dx \].

Step 3: Evaluate the Boundary Terms

Calculate the boundary term \( \left[ \frac{x^2}{2} \arcsin(x) \right]_0^{1/\sqrt{2}} \). When \( x = 0 \), the term is 0. When \( x = \frac{1}{\sqrt{2}} \), we have \( \frac{1/2}{2} \cdot \frac{\pi}{4} = \frac{\pi}{16} \).

Step 4: Simplify the Integral

We need to simplify \( \int_{0}^{1/\sqrt{2}} \frac{x^2}{2\sqrt{1-x^2}} \, dx \) further. Divide by the constant \(2\), it becomes \( \frac{1}{2} \int_{0}^{1/\sqrt{2}} \frac{x^2}{\sqrt{1-x^2}} \, dx \).

Step 5: Trigonometric Substitution

Perform the substitution \( x = \sin(\theta) \), with \( dx = \cos(\theta) \, d\theta \). The limits change from \( x = 0 \) to \( \theta = 0 \) and from \( x = \frac{1}{\sqrt{2}} \) to \( \theta = \frac{\pi}{4} \). The integral becomes: \[ \frac{1}{2} \int_{0}^{\pi/4} \frac{\sin^2(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \frac{1}{2} \int_{0}^{\pi/4} \sin^2(\theta) \, d\theta \].

Step 6: Use Identity to Integrate

Utilize the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \), changing the integral to \( \frac{1}{2} \int_{0}^{\pi/4} \frac{1 - \cos(2\theta)}{2} \, d\theta \).

Step 7: Compute Resulting Integrals

Calculate \( \frac{1}{4} \int_{0}^{\pi/4} (1 - \cos(2\theta)) \, d\theta \). This gives: \( \frac{1}{4} \left[ \theta - \frac{1}{2}\sin(2\theta) \right]_0^{\pi/4} = \frac{1}{4} \left[ \frac{\pi}{4} - 0 \right] = \frac{\pi}{16} \).

Step 8: Combine Results

The solution to the original integral is: \( \frac{\pi}{16} - \frac{\pi}{16} = 0 \).

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a valuable technique in calculus used to simplify integrals containing square roots. Typically, it involves substituting a trigonometric identity to transform the integrand into a form that is easier to manage. When you encounter terms like \( \sqrt{1 - x^2} \), substitution with trigonometric functions becomes handy.

For the integral of \( x \arcsin(x) \), the trigonometric substitution \( x = \sin(\theta) \) is utilized. Here's why:

• The substitution \( x = \sin(\theta) \) implies \( dx = \cos(\theta) \, d\theta \), which means the differential changes to match the function's transformation.
• The range of \( x \), from 0 to \( \frac{1}{\sqrt{2}} \), is mapped into an angle \( \theta \) from 0 to \( \frac{\pi}{4} \). This ensures that all variables align with the trigonometric identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \).

By making such a substitution, the integral \( \int \frac{x^2}{\sqrt{1-x^2}} \, dx \) transforms into an expression involving only trigonometric functions, thereby simplifying the process of integration.

Arcsin Function

The arcsin function, also known as the inverse sine function, is a common mathematical function denoted as \( \arcsin(x) \). It serves to "undo" the effect of the sine function by returning the angle whose sine is \( x \).

In calculus, the arcsin function is essential whenever we need to reverse or approach functions originally expressed through sine. For example:

• The derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \). This derivative plays a crucial role while applying integration by parts, as it defines the part of the function to be differentiated.
• The arcsin function is only defined for \( -1 \leq x \leq 1 \), meaning that any integral involving arcsin has to respect these boundaries.

In the given integral problem, \( u \) is set to \( \arcsin(x) \), supporting the need to calculate \( du \) effectively during integration by parts. This step links the original expression to a more solvable form.

Definite Integral

A definite integral applies the principles of integration to calculate the area under a curve between two specific bounds. This "boundaries" aspect distinguishes it from the indefinite integral, which eschews limits and typically involves a general antiderivative, often accompanied by a constant.

With definite integrals, mathematical evaluations undergo a refined procedure as follows:

• First, one computes the antiderivative of the integrand.
• Next, the antiderivative is evaluated at the upper and lower limits of the integral, respectively.
• The difference of these evaluations gives the final answer, representing the area under the curve between the specific points.

In the problem at hand, the integration process is completed by evaluating boundaries \( \left[ \frac{x^2}{2} \arcsin(x) \right]_0^{1/\sqrt{2}} \). Each bound is crucial in rendering the final numerical result, impacting the understanding of physical properties, such as total area under a curve, across the specified domain.