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Chapter 5: Q. 27 (page 417)

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�xcsc2x2dx

Short Answer

Expert verified

The solution of the given integral is 鈭�xcsc2x2dx=-12cotx2+C.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

鈭�xcsc2x2dx

02

Step 2. Solving the given integral using substitution method. 

Let

u=x2dudx=2xdu=2xdx12du=xdx

03

Step 3. This substitution changes the integral into 

鈭�xcsc2x2dx=12鈭�csc2udu鈭�xcsc2x2dx=12-cotu+C鈭�xcsc2x2dx=12-cotx2+C鈭�xcsc2x2dx=-12cotx2+C

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Most popular questions from this chapter

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

鈭�xx2-1dx

Show by differentiating (and then using algebra) that 鈭�cot鈦�sin鈭�1鈦�xand 鈭�1鈭�x2xare both antiderivatives of 1x21鈭�x2. How can these two very different-looking functions be an antiderivative of the same function?

Find three integrals in Exercises 27鈥�70 for which a good strategy is to use integration by parts with u=xand dv the remaining part.

Solve 鈭�x+2x2+4x32dxthe following two ways:

(a) with the substitution u=x2+4x;

(b) by completing the square and then applying the trigonometric substitution x + 2 = 2 sec u.

Why doesn鈥檛 the definite integral鈭�231-x2dx make sense? (Hint: Think about domains.)
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