Problem 3 - Calculus 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

$$ \int \frac{d x}{a^{2}-x^{2}}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right| \text { or } \frac{1}{a} \tan h^{-1}\left(\frac{x}{a}\right) $$

Step-by-Step Solution

Answer

\( \int \frac{dx}{a^2-x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \) or \( \frac{1}{a} \tanh^{-1} \left( \frac{x}{a} \right) + C \).

Step 1: Recognize Integral Type

The integral \( \int \frac{1}{a^2 - x^2} \, dx \) is one of the standard integral forms. It is related to the integrals involving inverse hyperbolic functions and logarithmic expressions.

Step 2: Substitute for Simplification

For the form \( \int \frac{1}{a^2 - x^2} \, dx \), we recognize it resembles the form \( \int \frac{1}{a^2 - u^2} \, du \). We could make the substitution \( x = a \sinh u \) or use partial fraction decomposition.

Step 3: Partial Fraction Decomposition

Express the integrand using partial fractions: \( \frac{1}{a^2-x^2} = \frac{1}{2a} \left( \frac{1}{a-x} - \frac{1}{a+x} \right) \). This decomposition splits the integral into two simpler parts.

Step 4: Integrate Each Fraction

Integrate each part separately: \[ \int \frac{1}{a-x} \, dx = -\log |a-x| + C_1 \] and \[ \int \frac{1}{a+x} \, dx = \log |a+x| + C_2 \].

Step 5: Combine Logarithmic Terms

Combine the results to get the full expression: \( \frac{1}{2a} (-\log |a-x| + \log |a+x|) = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \).

Step 6: Verify with Hyperbolic Form

Alternatively, verify the solution using \( \frac{1}{a} \tanh^{-1} \left( \frac{x}{a} \right) \). Recall the hyperbolic identity \( \tanh^{-1}(u) = \frac{1}{2} \log \left( \frac{1+u}{1-u} \right) \) and substitute \( u = \frac{x}{a} \).

Step 7: Final Solution

The integral solution is thus, either \( \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right| + C \) or \( \frac{1}{a} \tanh^{-1} \left( \frac{x}{a} \right) + C \) based on the form used.

Key Concepts

Partial Fraction Decomposition

Partial fraction decomposition is a technique used to break down complex rational expressions into simpler ones that are easier to integrate. When you have the integral \[ \int \frac{1}{a^2-x^2} \, dx \],you can apply partial fraction decomposition to express it as a sum of simpler fractions.

In this scenario, we decompose \( \frac{1}{a^2-x^2} \) into:

• \( \frac{1}{2a} \left( \frac{1}{a-x} - \frac{1}{a+x} \right) \).

This decomposition allows us to split the integral into two separate, simpler integrals, \[ \int \frac{1}{a-x} \, dx \]and \[ \int \frac{1}{a+x} \, dx \].
A key benefit of partial fraction decomposition is that it converts a complicated fraction into manageable terms without altering the original expression's value. This makes the integration process more straightforward by simplifying the task into dealing with basic logarithmic functions.

Inverse Hyperbolic Functions

Inverse hyperbolic functions are related to logarithmic functions and often appear in integrals where the integrand takes a specific form.In our exercise, the inverse hyperbolic function of interest is \( \tanh^{-1}(u) \).

Recall:

• \( \tanh^{-1}(u) = \frac{1}{2} \log \left( \frac{1+u}{1-u} \right) \).

In this integral, by recognizing \( u = \frac{x}{a} \), we successfully transformed our original integral into the equivalent hyperbolic form:

• \( \frac{1}{a} \tanh^{-1} \left( \frac{x}{a} \right) + C \).

Understanding inverse hyperbolic functions is essential because they offer alternative solutions and can simplify the calculation of integrals where substitutions are made.

Substitution Method

The substitution method is a powerful technique used to simplify integration problems by changing variables.
A common substitution approach for integrals resembling \( \int \frac{1}{a^2-x^2} \, dx \) is using the identity \( x = a \sinh(u) \).

The idea is to transform the integral into a simpler form:

• \( dx = a \cosh(u) \ du \).

This substitution transforms complicated variable expressions into linear or separable terms, making integration feasible.
The substitution method is beneficial in cases where initial inspection shows resemblance to a known integral form, facilitating quick and efficient solutions.

Logarithmic Expressions

Logarithmic expressions are crucial in the integration of rational functions, especially after applying partial fraction decomposition.
When we decomposed the fraction \( \frac{1}{a^2-x^2} \) into simpler parts, it resulted in two natural logarithm integrals:

• \( \int \frac{1}{a-x} \, dx = -\log |a-x| + C_1 \)
• \( \int \frac{1}{a+x} \, dx = \log |a+x| + C_2 \)

Logarithmic expressions allow us to manage functions with variables in both the numerator and denominator, effectively handling ratios and exponents.
Understanding how to integrate expressions of form \( \frac{1}{x} \) stems from the properties of logarithms.
This helps solve integrals that may seem complex but unravel naturally into logarithmic identities.