Problem 54 - Calculus Single Variable

Question

Calculate the given integral. $$ \int \frac{2 x^{2}}{\sqrt{x^{2}-1}} d x $$

Step-by-Step Solution

Answer

\( \int \frac{2x^2}{\sqrt{x^2-1}} \, dx = 2 \left(x \sqrt{x^2-1} - \ln|x+\sqrt{x^2-1}|\right) + C \).

Step 1: Identify the Integral Type

The given integral is \( \int \frac{2x^2}{\sqrt{x^2 - 1}} \, dx \). Notice that the integral resembles a function involving \( \sqrt{x^2 - 1} \), suggesting the use of trigonometric substitution, specifically with secant.

Step 2: Choose an Appropriate Trigonometric Substitution

Use the substitution \( x = \sec \theta \), where \( dx = \sec \theta \tan \theta \, d\theta \). Note that \( \sqrt{x^2 - 1} = \tan \theta \) in this substitution.

Step 3: Substitute and Simplify in Terms of \( \theta \)

Replace \( x \) and \( dx \) in the integral:\[ \int \frac{2(\sec \theta)^2}{\tan \theta} \sec \theta \tan \theta \, d\theta = \int 2 \sec^3 \theta \, d\theta. \]

Step 4: Simplify the Integral

The integral simplifies to \( \int 2\sec^3 \theta \, d\theta \). Use the identity \( \sec^3 \theta = \sec \theta(\sec^2 \theta) \) and integration by parts to solve this.

Step 5: Integration by Parts

Choose \( u = \sec \theta \) and \( dv = \sec^2 \theta \, d\theta \) so that \( du = \sec \theta \tan \theta \, d\theta \) and \( v = \tan \theta \). Integrating by parts, \[ \int \sec \theta \sec^2 \theta \, d\theta = \sec \theta \tan \theta - \int \sec \theta \tan^2 \theta \, d\theta. \]

Step 6: Simplify the Remaining Integral

Notice that \( \tan^2 \theta = \sec^2 \theta - 1 \). Substitute and simplify:\[ \int \sec \theta(\sec^2 \theta - 1) \, d\theta = \int \sec^3 \theta \, d\theta - \int \sec \theta \, d\theta. \] The second integral, \( \int \sec \theta \, d\theta \), is a standard result.

Step 7: Solve the Integral

The result is: \[ \int \sec^3 \theta \, d\theta = \frac{1}{2}(\sec \theta \tan \theta + \ln|\sec \theta + \tan \theta|) + C \] Combine with substitution to simplify to:\[ 2 \left(\sec \theta \tan \theta - \ln|\sec \theta + \tan \theta|\right) + C. \]

Step 8: Substitution Back to \( x \)

Revert the trigonometric identity back to the original variable \( x \). Remember that \( x = \sec \theta \) and \( \tan \theta = \sqrt{x^2 - 1} \), thus ensuring consistency with the original limits or expressions in terms of \( x \).

Step 9: Final Simplified Result

The integral simplifies to:\[ \int \frac{2x^2}{\sqrt{x^2 - 1}} \, dx = 2 \left(x \sqrt{x^2-1} - \ln|x + \sqrt{x^2-1}|\right) + C. \]

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a powerful technique in integral calculus that simplifies integrals involving square roots of quadratic expressions. When faced with an integral like \( \int \frac{2x^2}{\sqrt{x^2 - 1}} \, dx \), trigonometric substitution can help make the expression easier to integrate by transforming it into a trigonometric form.
Here's how it works: By using the substitution \( x = \sec \theta \), we can exploit the identity \( \sec^2 \theta - 1 = \tan^2 \theta \). This transforms the integral, replacing \( \sqrt{x^2 - 1} \) with \( \tan \theta \), which is much simpler to handle.
The steps are as follows:

• Choose a trigonometric substitution that matches the quadratic form under the square root. For expressions like \( \sqrt{x^2 - 1} \), \( x = \sec \theta \) is optimal.
• Differentiate the substitution to find \( dx \) in terms of \( d\theta \). Here, \( dx = \sec \theta \tan \theta \, d\theta \).
• Substitute both \( x \) and \( dx \) in the integral, converting all terms to trigonometric functions.
• Simplify the new integral as much as possible before proceeding with further integration techniques.

By simplifying the square root and converting the integral to a trigonometric form, we can more easily apply other methods like integration by parts.

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation and is used to integrate products of functions, typically when other methods are not directly applicable. In the context of this exercise, after applying trigonometric substitution, the integral becomes \( \int 2 \sec^3 \theta \, d\theta \). This is where integration by parts comes into play.
The formula for integration by parts is: \[\int u \, dv = uv - \int v \, du\]To apply this formula, select parts of the integral to act as \( u \) and \( dv \):

• Typically, choose \( u \) to be the function that becomes simpler when differentiated, and \( dv \) to be the part that remains manageable when integrated.
• In our example, let \( u = \sec \theta \) and \( dv = \sec^2 \theta \, d\theta \), thus \( du = \sec \theta \tan \theta \, d\theta \) and \( v = \tan \theta \).
• Substitute these into the formula to transform the integral and solve the resulting simpler integral.

Through this method, the complex trigonometric integral is broken down into simpler parts that can be integrated individually, allowing us to solve previously intractable integrals.

Antiderivative

An antiderivative, or indefinite integral, is a function whose derivative is the given function. Finding the antiderivative is the reverse process of differentiation and is central to solving integrals. When we compute \( \int \frac{2x^2}{\sqrt{x^2 - 1}} \, dx \), our goal is ultimately to find the antiderivative of the integrand.
The process involves several steps:

• Recognize the form of the integral to decide on an appropriate method, such as trigonometric substitution.
• Simplify the integrand using substitutions and other calculus techniques like integration by parts.
• Once the antiderivative is found, always remember to add the constant of integration \( C \) since there are infinite possible antiderivatives differing by a constant.

After employing trigonometric substitution and integration by parts, we find the antiderivative in trigonometric terms, which is then converted back to the original variable using the inverse of our substitution.
The final result incorporates all transformations, simplified back into the context of the original integrand, and includes the constant of integration to reflect all possible solutions.