Problem 25 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-28, perform the indicated integrations. \(\int \tan ^{-3} x \sec ^{4} x d x\)

Step-by-Step Solution

Answer

The integral is \(-\frac{1}{2}\cot^2 x + 2 \ln |\tan x| + \frac{1}{2}\tan^2 x + C\).

Step 1: Recognize Trigonometric Identities

To solve the integral \( \int \tan^{-3}x \sec^{4}x \, dx \), we first recognize trigonometric identities. We know that \( \sec^2 x = 1 + \tan^2 x \) and this can be useful for integration problems involving both secant and tangent. Also, remember that the derivative of \( \tan x \) is \( \sec^2 x \).

Step 2: Rewrite the Integral

Rewrite \( \sec^4 x \) using the identity \( \sec^2 x = 1 + \tan^2 x \). This makes \( \sec^4 x = (\sec^2 x)^2 = (1+\tan^2 x)^2 \). Substitute this back into the integral:\[ \int \tan^{-3}x \sec^{4}x \, dx = \int \tan^{-3}x (1+\tan^2x)^2 \sec^2 x \, dx \].

Step 3: Use Substitution

Let \( u = \tan x \). Then the differential \( du = \sec^2 x \, dx \). Substitute these into the integral:\[ \int u^{-3}(1+u^2)^2 \, du \]. This simplifies the problem considerably since we removed trigonometric functions.

Step 4: Expand and Integrate

Expand \((1+u^2)^2\):\((1 + u^2)^2 = 1 + 2u^2 + u^4\). Substituting back, we get:\[ \int (u^{-3})(1 + 2u^2 + u^4) \, du = \int (u^{-3} + 2u^{-1} + u) \, du \].

Step 5: Integrate Each Term

Perform the integration for each term separately:- \( \int u^{-3} \, du = -\frac{1}{2}u^{-2} + C_1 \)- \( \int 2u^{-1} \, du = 2 \ln |u| + C_2 \)- \( \int u \, du = \frac{1}{2}u^2 + C_3 \)Combine these results:\[ -\frac{1}{2}u^{-2} + 2 \ln |u| + \frac{1}{2}u^2 + C \].

Step 6: Substitute Back Into Original Variable

Replace \( u = \tan x \) back into the expression:\[ -\frac{1}{2}\frac{1}{\tan^2 x} + 2 \ln |\tan x| + \frac{1}{2}\tan^2 x + C \].

Step 7: Simplify the Expression

The expression can be simplified further as:- \(-\frac{1}{2}\frac{1}{\tan^2 x} = -\frac{1}{2}\cot^2 x \), where \( \cot x = \frac{1}{\tan x} \).So, the final simplified expression is:\[-\frac{1}{2}\cot^2 x + 2 \ln |\tan x| + \frac{1}{2}\tan^2 x + C\].

Key Concepts

Trigonometric Identities

Trigonometric identities are like secret codes in mathematics that help solve complex equations by simplifying expressions. In integration, these identities can transform complicated trigonometric functions into more manageable forms.

For instance, knowing that \( \sec^2 x = 1 + \tan^2 x \) can convert high power secant expressions into simple tangents. Such a conversion is quite handy when dealing with integrals involving both secant and tangent, like in our given exercise.

• The identity \( \sec^2 x = 1 + \tan^2 x \) simplifies integration problems involving both \( \sec x \) and \( \tan x \).
• Recognize derivatives: The derivative of \( \tan x \) is \( \sec^2 x \), suggesting potential substitution paths.

Understanding these identities enables you to make strategic substitutions, leading to more straightforward solutions.

Substitution Method

The substitution method is like finding a hidden shortcut in math. It transforms a too-complicated expression into something much simpler. This method works wonders for integrals by replacing a complicated function with a simpler one using a new variable.

In our problem, we use substitution to remove the trigonometric functions and simplify the integral. By letting \( u = \tan x \), we can take advantage of the fact that \( du = \sec^2 x \, dx \).

• This substitution turns the integral \( \int \tan^{-3}x \sec^{4}x \, dx \) into \( \int u^{-3}(1+u^2)^2 \, du \).
• This new form is free from trigonometric terms, making the integral easier to solve.

The goal of substitution is to navigate around complexity using something you already know.

Integration of Powers

Integration of powers is a fundamental technique in calculus, especially when you're simplifying expressions into polynomials. This allows us to tackle each term separately in integrals of the form \( \int x^n \, dx \).

For the exercise at hand, after substitution, we expand and integrate simple power terms. Consider the polynomial expansion \((1 + u^2)^2 = 1 + 2u^2 + u^4\). Each term can then be integrated independently:

• \( \int u^{-3} \, du = -\frac{1}{2}u^{-2} + C_1 \)
• \( \int 2u^{-1} \, du = 2 \ln |u| + C_2 \)
• \( \int u \, du = \frac{1}{2}u^2 + C_3 \)

You combine these terms to formulate the full solution. Integrating powers separately helps you focus on smaller, manageable problems within the broader integral challenge.