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

∫-ππsinxcosxdx

Short Answer

Expert verified

The solution of the given integral is ∫-ππsinxcosxdx=0.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

∫-ππsinxcosxdx

02

Step 2. Using the substitution method. 

Let

u=sinxdudx=cosxdu=cosxdx

03

Step 3. Using the information in equations, we can change variables completely:

∫-ππsinxcosxdx=∫x=-Ï€x=Ï€udu∫-ππsinxcosxdx=u1+11+1x=-Ï€x=π∫-ππsinxcosxdx=u22x=-Ï€x=π∫-ππsinxcosxdx=sin2x2x=-Ï€x=π∫-ππsinxcosxdx=12(sinx)2x=-Ï€x=π∫-ππsinxcosxdx=12(²õ¾±²ÔÏ€)2-{sin(-Ï€)}2∫-ππsinxcosxdx=12(²õ¾±²ÔÏ€)2-{-sin(Ï€)}2∫-ππsinxcosxdx=12(0)2-(0)2∫-ππsinxcosxdx=0

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Most popular questions from this chapter

Suppose you use polynomial long division to divide p(x) by q(x), and after doing your calculations you end up with the polynomial x2-x+3 as the quotient above the top line, and the polynomial 3x − 1 at the bottom as the remainder. Thenp(x)=___andp(x)q(x)=____

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving∫1x2−4dx.

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving∫1x2−4dx.

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving∫1x2+4dx.

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving∫x2+4−5/2dx

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form x2−a2.

(f) True or False: Trigonometric substitution doesn’t solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with x=asinu, we must consider the cases x>a and x<-a separately.

(h) True or False: When using trigonometric substitution with x=asecu, we must consider the cases x>a and x<-a separately.

Solve the integral:∫x2cosxdx.

Find three integrals in Exercises 21–70 that we can anti-differentiate immediately after algebraic simplification.

Complete the square for each quadratic in Exercises 28–33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

x2+6x−2

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