Problem 1 - Calculus 2 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

$$ \int \frac{d t}{\sqrt{(t+\alpha)(t+\beta)}}=2 \log [\sqrt{t+\alpha}+\sqrt{t+\beta}] $$

Step-by-Step Solution

Answer

Use substitution and direct integration techniques to solve the integral.

Step 1: Understand the Integral Expression

The given integral is \( \int \frac{d t}{\sqrt{(t+\alpha)(t+\beta)}} \). This is a type of integral that can often be solved by substitution or direct integration techniques.

Step 2: Use Substitution

A useful substitution for this integral type is to let \( x = \sqrt{t+\alpha} + \sqrt{t+\beta} \). This implies \( \sqrt{t+\alpha} = \frac{x^2 - (t+\beta)}{2x} \) and \( \sqrt{t+\beta} = \frac{x^2 - (t+\alpha)}{2x} \). The differential is derived from this.

Step 3: Simplify the Expression

Compute the differential using the chain rule and adjust the original integral to reflect the new variables. When solving, aim to transform the integral into a more standard form that can be evaluated directly.

Step 4: Integrate and Solve

After substitution and simplification, the integral turns into an expression that can be integrated using known techniques. This includes direct integration resulting in \( 2 \log[\sqrt{t+\alpha}+\sqrt{t+\beta}] + C \) as the upper level solution.

Key Concepts

Integration by Substitution

Integration by substitution is a powerful technique used to simplify complex integrals by changing variables. This is similar to the chain rule in differentiation where we "substitute" a part of the integral with a single variable. In our original exercise, we apply substitution to tackle the integral \( \int \frac{d t}{\sqrt{(t+\alpha)(t+\beta)}} \). Here, a strategic substitution is chosen: \( x = \sqrt{t+\alpha} + \sqrt{t+\beta} \). This is done to simplify the expression under the square root, allowing us to transform it into a more manageable form.

• Identify part of the integrand to substitute with a new variable \( x \).
• Express \( t \) in terms of \( x \).
• Find the differential \( dt \) in terms of \( dx \) using the chain rule.

This process ultimately leads to an integral that is easier to evaluate. The substitution allows integrating functions that look initially unsolvable, by transforming them into expressions that are already well-known.

Chain Rule in Integration

The chain rule is a fundamental concept in calculus, mainly used in differentiation. However, in integration, it primarily shows up through substitution, wherein we adjust the differential as per the change of variable.In our example, once the substitution \( x = \sqrt{t+\alpha} + \sqrt{t+\beta} \) is made, we compute the differential \( dx \).

• This involves differentiating \( x \) with respect to \( t \).
• The resulting differential allows us to replace \( dt \) in the integral with terms in \( dx \).

The chain rule here ensures that any change in the variable of integration correctly accounts for the original variables' relation. This careful handling of differentials ensures that the transformation remains valid throughout the process, and helps in reducing the original integral into a new form that is easier to solve.

Logarithmic Integration

Logarithmic integration refers to an integration process where the result involves logarithms. Often occurring when integrating functions similar to \( \frac{1}{x} \), it's essential in expressing the final solution in terms of log functions.In the given integral \( \int \frac{d t}{\sqrt{(t+\alpha)(t+\beta)}} \), following the substitution and simplification, we arrive at a form that yields a logarithmic result.

• The integration results in \( 2 \log[\sqrt{t+\alpha}+\sqrt{t+\beta}] \).
• This expression denotes the logarithmic representation necessary to capture the behavior of the integral across its domain.

Logarithmic forms often arise when dealing with radicals and fractions, thanks to the inverse operations they represent. Recognizing when and how to work with logarithmic integrals allows for precise solutions in examples like this, where directly solving at first glance would seem complicated.