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Chapter 5: Q 35. (page 452)

Solve the following integral.
$\int c s c^{4} x d x$

Short Answer

Expert verified

Answer is $\begin{array}{rcl} \frac{- c o t^{3} x}{3} - c o t x + C \end{array}$

Step by step solution

01

Step 1. Given information

An integral is $\int c s c^{4} x d x$

02

Step 2. Explanation

$\begin{array}{rcl} \int c s c^{4} x d x & = & \int c s c^{2} x c s c^{2} x d x \\ = & \int (1 + c o t^{2} x) c s c^{2} x d x \end{array}$

Let $u = c o t x$

$\begin{array}{rcl} \int c s c^{4} x d x & = & - \int u^{2} + 1 d u \\ = & \frac{- u^{3}}{3} - u + C \\ = & \frac{- c o t^{3} x}{3} - c o t x + C \end{array}$

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Most popular questions from this chapter

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int \frac{\sqrt{9 - x^{2}}}{x} d x$

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{x^{2} - 4} d x$ .

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving $\int \frac{1}{\sqrt{x^{2} - 4}} d x$ .

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving $\int \frac{1}{x^{2} + 4} d x .$

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving $\int {(x^{2} + 4)}^{- 5 / 2} d x$

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form $\sqrt{x^{2} - a^{2}}$ .

(f) True or False: Trigonometric substitution doesn’t solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with $x = a \sin u$ , we must consider the cases $x > a$ and $x < - a$ separately.

(h) True or False: When using trigonometric substitution with $x = a s e c u$ , we must consider the cases $x > a$ and $x < - a$ separately.

Consider the integral $\int \frac{1}{x^{2} \sqrt{1 - x^{2}}} d x$ from the reading at the beginning of the section.

(a) Use the inverse trigonometric substitution $u = \sin^{- 1} x$ to solve this integral.

(b) Use the trigonometric substitution $x = \sin u$ to solve the integral.

(c) Compare and contrast the two methods used in parts (a) and (b).

Find three integrals in Exercises 21–70 in which the denominator of the integrand is a good choice for a substitution u(x).

Solve each of the integrals in Exercises 39–74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$\int \frac{x}{\sqrt{x^{2} - 1}} d x$

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