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Chapter 5: Q 32. (page 452)

Solve the following integral.
$\int s e c^{6} x d x$

Short Answer

Expert verified

Answer is $\frac{\tan^{5} x}{5} + \frac{2 \tan^{3} x}{3} + \tan x + C$

Step by step solution

01

Step 1. Given information

An integral is $\int s e c^{6} x d x$

02

Step 2. Explanation

$\begin{array}{rcl} \int s e c^{6} x d x & = & \int s e c^{2} x s e c^{4} x d x \\ = & \int {(\tan^{2} x + 1)}^{2} s e c^{2} x d x \end{array}$

Let $\tan x = u$

$\begin{array}{rcl} \int s e c^{6} x d x & = & \int {(u^{2} + 1)}^{2} d u \\ = & \int u^{4} + 2 u^{2} + 1 d u \\ = & \frac{u^{5}}{5} + \frac{2 u^{3}}{3} + u + C \\ = & \frac{\tan^{5} x}{5} + \frac{2 \tan^{3} x}{3} + \tan x + C \end{array}$

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Most popular questions from this chapter

Show by differentiating (and then using algebra) that $- \cot (\sin^{- 1} x)$ and $- \frac{\sqrt{1 - x^{2}}}{x}$ are both antiderivatives of $\frac{1}{x^{2} \sqrt{1 - x^{2}}}$ . How can these two very different-looking functions be an antiderivative of the same function?

Which of the integrals that follow would be good candidates for trigonometric substitution? If a trigonometric substitution is a good strategy, name the substitution. If another method is a better strategy, explain that method.

$(a) \int \frac{4 + x^{2}}{x} d x$ $(b) \int \frac{x}{4 + x^{2}} d x$

role="math" localid="1648759296940" $(c) \int \frac{x^{2}}{4 + x^{2}} d x$ $(d) \int \frac{16 - x^{4}}{4 + x^{2}} d x$

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).

Explain why, if $x = a \sin u$ , then $\sqrt{a^{2} - x^{2}} = a \cos u$ . Your explanation should include a discussion of domains and absolute values.

Solve the integral: $\int \frac{x^{2}}{e^{x}} d x$

See all solutions

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