Problem 35 - Calculus Single Variable

Question

Use the method of substitution to evaluate the definite integrals. $$ \int_{0}^{\pi / 4} \frac{24 \tan (x) \sec ^{2}(x)}{(1+2 \tan (x))^{2}} d x $$

Step-by-Step Solution

Answer

The value of the definite integral is \( 6 \ln 3 - 4 \).

Step 1: Identify the Substitution

To solve the integral \( \int_{0}^{\pi / 4} \frac{24 \tan(x) \sec^2(x)}{(1+2 \tan(x))^{2}} \, dx \), we begin by identifying a suitable substitution. Notice that the derivative of \( \tan(x) \) is \( \sec^2(x) \), so a natural substitution would be \( u = 1 + 2\tan(x) \). This simplifies the expression inside the integrand.

Step 2: Differentiate the Substitution

Differentiate both sides of the substitution \( u = 1 + 2\tan(x) \) with respect to \( x \) to find \( du \): \[ du = 2\sec^2(x) \, dx \]Thus, \( dx = \frac{du}{2\sec^2(x)} \).

Step 3: Rewrite the Integral

Substitute \( u \) and \( dx \) into the integral to transform it:\[ \int \frac{24 \tan(x) \sec^2(x)}{u^{2}} \, \frac{du}{2\sec^2(x)} \]Simplify the expression:\[ \int \frac{12 \tan(x)}{u^{2}} \, du \]

Step 4: Express \( \tan(x) \) in Terms of \( u \)

Since \( u = 1 + 2\tan(x) \), solve for \( \tan(x) \):\[ \tan(x) = \frac{u - 1}{2} \]Substitute this back into the integral:\[ \int \frac{12 (u - 1)}{2u^{2}} \, du = \int \frac{6(u - 1)}{u^{2}} \, du \]

Step 5: Simplify the Integral

Separate the fraction and simplify:\[ \int \left(\frac{6}{u} - \frac{6}{u^2}\right) \, du = 6 \int \frac{1}{u} \, du - 6 \int \frac{1}{u^2} \, du \]

Step 6: Integrate Each Term

Evaluate each integral separately:- \( 6 \int \frac{1}{u} \, du = 6 \ln|u| + C \)- \( -6 \int \frac{1}{u^2} \, du = 6 \cdot \frac{1}{u} + C \)Thus, the indefinite integral becomes \[ 6 \ln|u| + \frac{6}{u} + C \].

Step 7: Apply Limits

Apply the original limits by converting them from \( x \) to \( u \) values:- When \( x = 0 \), \( u = 1 + 2 \tan(0) = 1 \).- When \( x = \frac{\pi}{4} \), \( u = 1 + 2 \tan\left(\frac{\pi}{4}\right) = 3 \).Substitute these into the definite integral:\[ \left[ 6 \ln|u| + \frac{6}{u} \right]_{1}^{3} \]

Step 8: Calculate the Definite Integral

Compute the definite integral:- \( \left. 6 \ln|u| + \frac{6}{u} \right|_{1}^{3} = \left(6 \ln 3 + \frac{6}{3}\right) - \left(6 \ln 1 + \frac{6}{1}\right) \)- Simplify: \( 6 \ln 3 + 2 - 0 - 6 = 6 \ln 3 - 4 \).

Key Concepts

Method of Substitution

The method of substitution is a powerful technique used in integral calculus to simplify the process of evaluating integrals. It is akin to the reverse process of the chain rule used in differentiation. This technique is particularly useful when the integral involves a function and its derivative. In our problem, we encounter a complex integrand

• \( \int_{0}^{\pi / 4} \frac{24 \tan (x) \sec ^{2}(x)}{(1+2 \tan (x))^{2}} \, dx \),
• which can be simplified using substitution.

To apply the method of substitution, identify a part of the integrand that can be set as a new variable, often denoted as \( u \). In this case, we choose \( u = 1 + 2\tan(x) \) because its derivative \( d(1+2\tan(x))/dx = 2\sec^2(x) \) is part of our integral. This relationship allows us to express \( dx \) in terms of \( du \), thus simplifying the integral into a more manageable form. By turning \( dx = \frac{du}{2\sec^2(x)} \), we effectively rewrite and simplify the integral, making it easier to evaluate.

Trigonometric Functions

Trigonometric functions, such as \( \tan(x) \) and \( \sec(x) \), are frequently encountered in calculus. Understanding their properties is essential for solving integrals involving these functions. In the given integral,

• \( \tan(x) \) and \( \sec^2(x) \) are both critical to the substitution method.

The function \( \tan(x) \), short for tangent, represents the ratio of the opposite side to the adjacent side in a right triangle. Its derivative, \( \sec^2(x) \), or the square of the secant function, represents the square of the reciprocal of the cosine function.

• The secant function, \( \sec(x) = 1/\cos(x) \), plays a vital role when differentiating trigonometric functions.

In this exercise, recognizing that \( \frac{d}{dx} (\tan(x)) = \sec^2(x) \) allows us to make the substitution \( u = 1 + 2\tan(x) \) work seamlessly, simplifying our integral into a format suitable for straightforward integration.

Integral Calculus

Integral Calculus is a branch of mathematics focused on the accumulation of quantities and the areas under and between curves. The primary tool within this branch is the integral, which comes in two varieties: definite and indefinite. In this exercise, we deal with a definite integral

• \( \int_{0}^{\pi / 4} \frac{24 \tan (x) \sec ^{2}(x)}{(1+2 \tan (x))^{2}} \, dx \).

A key feature of definite integrals is the application of limits, which provide the bounds for integration and result in a specific numerical value. The goal is to transform the complex integrand into something simpler. Through substitution, we simplify the expressions and calculate the integral's value between the new limits. By converting our bounds \( x = 0 \) and \( x = \pi / 4 \) into \( u \) values, the integral becomes

• \( \int_{1}^{3} (\frac{6}{u} - \frac{6}{u^2}) \, du \).

After evaluating the simplified integral and applying these limits, the result is the specific accumulated value for the problem posed, which is found to be \( 6 \ln 3 - 4 \). Integral calculus, through methods such as substitution, makes the calculation of complex and seemingly intimidating integrals manageable.