Problem 65 - Calculus Single Variable

Question

Integrate by parts. This will result in an integrand of the form \(P(x) / Q(x)\) where \(P(x)\) and \(Q(x)\) are polynomials with the degree of \(P(x)\) greater than or equal to the degree of \(Q(x)\). Such an integrand is handled by performing polynomial division to put \(P(x) / Q(x)\) into the form \(r(x)+s(x) / Q(x)\) where \(r(x)\) and \(s(x)\) are polynomials with the degree of \(s(x)\) less than the degree of \(Q(x)\) $$ \int_{0}^{1} 2 x \arctan (x / \sqrt{3}) d x $$

Step-by-Step Solution

Answer

The evaluated integral is \( \sqrt{3} \arctan(1/\sqrt{3}) - 1 \).

Step 1: Integration by Parts Formula

The integration by parts formula is given by the integration rule \( \int u \, dv = uv - \int v \, du \). We apply this rule to solve the integral \( \int 2x \arctan(x/\sqrt{3}) \ dx \). We need to choose \( u \) and \( dv \).

Step 2: Choose u and dv

In the integral \( \int 2x \arctan(x/\sqrt{3}) \ dx \), choose \( u = \arctan(x/\sqrt{3}) \) because it simplifies upon differentiation, and \( dv = 2x \, dx \) because it simplifies upon integration.

Step 3: Differentiate u and Integrate dv

Differentiate \( u \) to find \( du \):\[ du = \frac{d}{dx} \left( \arctan(x/\sqrt{3}) \right) = \frac{1}{\sqrt{3}(1+(x/\sqrt{3})^2)} \, dx = \frac{1}{x^2+3} \, dx \]Integrate \( dv \) to find \( v \):\[ v = \int 2x \, dx = x^2 \]

Step 4: Apply Integration by Parts

Substitute \( u \), \( v \), \( du \), and \( dv \) into the integration by parts formula:\[ \int 2x \arctan(x/\sqrt{3}) \ dx = x^2 \arctan(x/\sqrt{3}) - \int x^2 \frac{1}{x^2+3} \, dx \].

Step 5: Simplify the Remaining Integral

Simplify the integral \( \int \frac{x^2}{x^2+3} \, dx \). Write \( \frac{x^2}{x^2+3} = 1 - \frac{3}{x^2+3} \), leading to:\[ \int \left(1 - \frac{3}{x^2+3}\right) \, dx = \int 1 \, dx - 3 \int \frac{1}{x^2+3} \, dx \].

Step 6: Integrate Each Term

Integrate term by term:\( \int 1 \, dx = x \) and \( \int \frac{1}{x^2+3} \, dx = \frac{1}{\sqrt{3}}\arctan\left( \frac{x}{\sqrt{3}} \right) \). Thus:\[ \int \frac{x^2}{x^2+3} \, dx = x - 3\left( \frac{1}{\sqrt{3}}\arctan\left( \frac{x}{\sqrt{3}} \right) \right) \].

Step 7: Combine the Results

Combine the results from integration by parts and simplify:\[ \int 2x \arctan(x/\sqrt{3}) \ dx = x^2 \arctan(x/\sqrt{3}) - \left( x - \sqrt{3} \arctan(x/\sqrt{3}) \right) \].

Step 8: Evaluate Definite Integral

Evaluate the definite integral from 0 to 1:- \( x=1 \): Substitute into the expression to get: \[ 1^2 \arctan(1/\sqrt{3}) - \left( 1 - \sqrt{3} \arctan(1/\sqrt{3}) \right) \].- \( x=0 \): Substitute into the expression to get zero.Simplifying these gives:\[ \int_{0}^{1} 2x \arctan(x/\sqrt{3}) \, dx = \sqrt{3} \arctan(1/\sqrt{3}) - 1 \].

Key Concepts

Polynomial Division

In mathematics, polynomial division is akin to the long division we perform with numbers, but it's applied to polynomials. When dividing two polynomials, say \( P(x) \) by \( Q(x) \), the goal is to express the division in the form \( r(x) + \frac{s(x)}{Q(x)} \) where \( r(x) \) and \( s(x) \) are polynomials.
Here, \( r(x) \) is the quotient and \( s(x) \) is the remainder. The degree of \( s(x) \) should always be less than that of \( Q(x) \).

To perform polynomial division, you repeatedly divide the leading term of the dividend by the leading term of the divisor to determine each term of the quotient. You subtract the result from the original polynomial and repeat the process with the remainder until the degree of the remainder is less than the divisor.

This method can simplify a complicated rational function, making it easier to integrate or differentiate. In problems like the one presented, where you need to integrate a fraction involving polynomials, simplifying the integrand through polynomial division is often a necessary step.

Definite Integrals

Definite integrals are a fundamental concept in calculus that represent the signed area under a curve on a specific interval. When evaluating a definite integral, you determine the accumulation of quantities, like area, between two points on the \(x\)-axis, typically denoted by the bounds \(a\) and \(b\).

The definite integral of a function \( f(x) \) from \( a \) to \( b \) is written as:
\[ \int_{a}^{b} f(x) \, dx \]

To calculate a definite integral, you find the antiderivative (or the indefinite integral) of the function and then apply the Fundamental Theorem of Calculus. This involves evaluating the antiderivative at the upper bound and subtracting its value at the lower bound.

In our exercise, after simplifying the integral using polynomial division and integration by parts, you compute the definite integral by substituting the bounds (from 0 to 1) into the expression obtained. These calculations help to find exact numerical results for integration problems.

Trigonometric Integrals

Trigonometric integrals involve functions of trigonometric identities, such as \( \sin(x) \), \( \cos(x) \), or, as in the given problem, \( \arctan(x) \). These types of integrals require special techniques for solving, often involving substitution or integration by parts.

In our example, \( \arctan(x/\sqrt{3}) \) appears in the integral. The derivative of the arctan function is crucial here:

• The derivative of \( \arctan(x) \) is \( \frac{1}{1+x^2} \), making it manageable when combined with other parts of the expression.
• When integrated, the resulting arctan provides a means to return to simpler forms, often simplifying to inverse trigonometric functions.

The key is recognizing when and how these trigonometric identities play out in integrals. Utilizing these strategies helps break down complex problems into elementary forms that are more readily solved, promoting a deeper understanding of the relationships between different functions within calculus.