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Chapter 8: Problem 49

Evaluate the following integrals. $$\int \tan ^{3} x \sec ^{9} x d x$$

Short Answer

Expert verified
Question: Evaluate the integral $$\int \tan ^{3} x \sec ^{9} x d x$$ Solution: Using integration by substitution and trigonometric identities, the evaluated integral is $$\int \tan ^{3} x \sec ^{9} x d x = \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^6 x + \frac{3}{8} \tan^8 x + \frac{1}{10} \tan^{10} x + C$$.

Step by step solution

01

Write down the given integral

The integral we need to evaluate is: $$\int \tan ^{3} x \sec ^{9} x d x$$
02

Perform integration by substitution

Let's perform a substitution. Let \(u = \tan x\) which implies \(du = \sec^2 xdx\). We'll also express the given integral in terms of \(u\): $$\int \tan ^{3} x \sec ^{9} x d x = \int u^3 (\sec^7 x) (\sec^2 x) d x = \int u^3 \sec^7 x d u$$ Now, we need to express \(\sec^7x\) in terms of \(u\). We use the identity, \(\sec^2 x = 1 + \tan^2 x = 1 + u^2\) and substitute it in our integral.
03

Replace the trigonometric function with the substitution

Substituting \(1 + u^2\) in place of \(\sec^2 x\): $$\int u^3 \sec^7 x d u = \int u^3 (1+u^2)^3 d u$$
04

Expand the polynomial and integrate term by term

Expand \((1 + u^2)^3\) and multiply with \(u^3\) to get a polynomial: $$\int u^3 (1+u^2)^3 d u = \int u^3 (1+3u^2+3u^4+u^6) d u = \int (u^3 + 3u^5 + 3u^7 + u^9) du$$ Now, integrate term by term: $$\int (u^3 + 3u^5 + 3u^7 + u^9) du = \frac{1}{4} u^4 + \frac{1}{2} u^6 + \frac{3}{8} u^8 + \frac{1}{10} u^{10} + C$$
05

Substitute back the original function

Replace \(u\) with \(\tan x\) to get the result in terms of \(x\): $$\frac{1}{4} u^4 + \frac{1}{2} u^6 + \frac{3}{8} u^8 + \frac{1}{10} u^{10} + C = \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^6 x + \frac{3}{8} \tan^8 x + \frac{1}{10} \tan^{10} x + C$$ So, the evaluated integral is: $$\int \tan ^{3} x \sec ^{9} x d x = \frac{1}{4} \tan^4 x + \frac{1}{2} \tan^6 x + \frac{3}{8} \tan^8 x + \frac{1}{10} \tan^{10} x + C$$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are fundamental in calculus, especially when dealing with integrals involving angles. They relate the angles of a triangle to the lengths of its sides. In our problem, we focus primarily on the two functions, \( \tan x \) and \( \sec x \).
  • The tangent function, \( \tan x \), represents the ratio of the opposite side to the adjacent side in a right-angled triangle.
  • The secant function, \( \sec x \), is the reciprocal of the cosine function, \( \sec x = \frac{1}{\cos x} \).
Understanding these functions and their properties helps us transform integrals into more manageable expressions. In this exercise, recognizing the identity \( \sec^2 x = 1 + \tan^2 x \) was pivotal. This identity facilitated the substitution process, allowing us to simplify the integration effort by expressing everything in terms of one trigonometric function. With practice, these transformations become more intuitive.
Substitution Method
The substitution method, also known as the \( u \)-substitution, is a powerful technique in calculus for simplifying integrals. This involves substituting part of the integral with a new variable to ease the integration process.We strategically chose \( u = \tan x \) because it simplified the integrand by directly using the identity \( \sec^2 x = 1 + \tan^2 x \). This transformed our integral from \( \int \tan^3 x \sec^9 x dx \) into \( \int u^3 (1+u^2)^3 du \).Steps to use substitution:
  • Identify a part of the integral to substitute, making the integral simpler.
  • Find the differential, \( du \), for the chosen \( u \).
  • Rewrite the integral in terms of \( u \) and \( du \).
  • Integrate with respect to \( u \).
  • Finally, substitute back the original expression for \( u \).
This method is especially helpful when encountering products or compositions of functions, making it a staple in solving complex integrals.
Polynomial Expansion
Polynomial expansion involves expanding expressions like \((a + b)^n\) into a sum of terms. This concept was critical in our integration solution when we had to integrate the expression \((1+u^2)^3\).Using the Binomial Theorem:
  • \((1+u^2)^3\) expands to \(1 + 3u^2 + 3u^4 + u^6\).
The key idea is to break down the complex expression into simpler polynomial terms, which are easier to integrate term-by-term.Steps:
  • Expand the polynomial expression.
  • Distribute the integration across each term.
  • Integrate each term individually using basic antiderivative rules. For example, the antiderivative of \(u^n\) is \(\frac{u^{n+1}}{n+1}\).
By expanding and simplifying even complex expressions, it becomes straightforward to apply direct integration rules to each term, leading to the final solution.

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Most popular questions from this chapter

For a real number \(a\), suppose \(\lim _{x \rightarrow a^{+}} f(x)=-\infty\) or \(\lim _{x \rightarrow a^{+}} f(x)=\infty .\) In these cases, the integral \(\int_{a}^{\infty} f(x) d x\) is improper for two reasons: \(\infty\) appears in the upper limit and \(f\) is unbounded at \(x=a .\) It can be shown that \(\int_{a}^{\infty} f(x) d x=\int_{a}^{c} f(x) d x+\int_{c}^{\infty} f(x) d x\) for any \(c>a .\) Use this result to evaluate the following improper integrals. $$\int_{1}^{\infty} \frac{d x}{x \sqrt{x-1}}$$

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