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Chapter 5: Q. 24 (page 496)

Use whatever method you like to solve each of the given definite and indefinite integrals. These integrals are neither in order of difficulty nor in order of technique. Many of the integrals can be solved in more than one way.

鈭�tanxlncosxdx

Short Answer

Expert verified

The result is-ln2cosx2+C.

Step by step solution

01

Step 1. Given information.

Consider the given integral,

鈭�tanxlncosxdx

02

Step 2. Solve the Integral.

Simplify and solve the integral.

鈭�tanxlncosxdx=鈭�sinxcosxlncosxdx=鈭�sinxcosxlncosxdxt=cosx,dt=-sinxdx=-鈭�1tlntdtu=lnt,du=1tdt=-鈭�udu=-u22+C=-lnt22+C=-lncost22+C

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Most popular questions from this chapter

Solve the integral :鈭�xex2+1dx

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.

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鈭�1x2鈭�4dx.

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

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

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

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

(f) True or False: Trigonometric substitution doesn鈥檛 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 given definite integral.

鈭�45鈥�1xx2+9dx

Describe two ways in which the long-division algorithm for polynomials is similar to the long-division algorithm for integers and then two ways in which the two algorithms are different.
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