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Chapter 5: Q 39. (page 452)

Solve the following integral.
$\int s e c x \tan^{3} x d x$

Short Answer

Expert verified

Answer is $\frac{s e c^{3} x}{3} - s e c x + C$

Step by step solution

01

Step 1. Given information

An integral is $\int s e c x \tan^{3} x d x$

02

Step 2. Explanation

$\int s e c x \tan^{3} x d x = \int (s e c^{2} x - 1) s e c x \tan x d x$

Let $u = s e c x$

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

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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.

$\int \frac{1}{{(x^{2} + 4)}^{\frac{3}{2}}} d x$

Solve the integral: $\int x^{2} e^{3 x} d x$ .

Solve the integral

$\int \sqrt{3 - x} \sqrt{x - 1} d x$

Find three integrals in Exercises 27–70 for which a good strategy is to use integration by parts with $u = x$ and dv the remaining part.

Why don’t we ever have cause to use the trigonometric substitution $x = \cos u$ ?

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