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Chapter 5: Q. 70 (page 418)

Solve each of the integrals in Exercises 21鈥�70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)

鈭�exe2x鈭�1dx

Short Answer

Expert verified

The solution of the given integral is 鈭�exe2x鈭�1dx=cos-1ex+C.

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

鈭�exe2x鈭�1dx

02

Step 2. Using the substitution method.

Let

u=exdudx=exdu=exdx

03

Step 3. This substitution changes the integral into 

鈭�exe2x鈭�1dx=鈭�ex(ex)2鈭�1dx鈭�exe2x鈭�1dx=鈭�1u2鈭�1du鈭�exe2x鈭�1dx=-鈭�11-u2du鈭�exe2x鈭�1dx=--cos-1u+C鈭�exe2x鈭�1dx=cos-1u+C鈭�exe2x鈭�1dx=cos-1ex+C

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Most popular questions from this chapter

Solve the integral鈭�x3x2-1dxthree ways:

(a) with the substitution u=x2-1,followed by back substitution;

(b) with integration by parts, choosing localid="1648814744993" u=x2anddv=xx2-1dx;

(c) with the trigonometric substitution x = sec u.

Write sin(2cos-1x)as an algebraic function.

For each function u(x) in Exercises 9鈥�12, write the differential du in terms of the differential dx.

u(x)=sinx

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

鈭�1udu

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

鈭�sinudu
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