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Chapter 5: Q 44. (page 429)

Solve the integral: $\int \frac{3 x + 1}{s e c x} d x$

Short Answer

Expert verified

The required answer is $(3 x + 1) \sin x + 3 \cos x + c$ .

Step by step solution

01

Step 1. Given information.

We have given integral is $\int \frac{3 x + 1}{s e c x} d x$ .

02

Step 2. Solve the integration by parts .

We have,

$u = 3 x + 1 d u = 3 d x$

and

$d v = \cos x d x v = \int \cos x d x v = \sin x$

The formula of integration by parts is $\int u d v = u v - \int v d u$ .

localid="1648742547412" $\int \frac{3 x + 1}{s e c x} d x = (3 x + 1) \sin x - \int 3 \sin x d x = (3 x + 1) \sin x - 3 \int \sin x d x = (3 x + 1) \sin x - 3 (- \cos x) + c = (3 x + 1) \sin x + 3 \cos x + c$

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Most popular questions from this chapter

Suppose v(x) is a function of x. Explain why the integral

of dv is equal to v (up to a constant).

Explain how to know when to use the trigonometric substitutions $x = a \sin u, x = a \tan u, a n d x = a \sec u$ , Describe the trigonometric identity and the triangle that will be needed in each case. What are the possible values for $x$ and $u$ in each case?

Find three integrals in Exercises 39–74 that can be solved by using a trigonometric substitution of the form $x = a \tan u$ .

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).

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.

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

See all solutions

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