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Solve each of the integrals in Exercises 27–70. Some integrals require integration by parts, and some do not. (The last two exercises involve hyperbolic functions.)

∫arcsin2xdx

Short Answer

Expert verified

The value of the integral isxarcsin(2x)+12-4x2+1.

Step by step solution

01

Step 1. Given information.

The given integral is∫arcsin2xdx.

02

Step 2. Substitution.

Now,

u=2x;du=2dx∫arcsin2xdx=12∫arcsin(u)duw=arcsinu;dw=du-u2+1v=u;dv=du

03

Step 3. Value of the integral.

The value of the integral is,

12∫arcsin(u)du=12(uarcsinu)-∫u-u2+1duv=-u2+1;dv=-2udu12(uarcsinu)-∫u-u2+1du=u2arcsinu-12∫-12vdv=u2arcsinu+14∫1vdv=u2arcsinu+14v1212=xarcsin(2x)+12-4x2+1

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Most popular questions from this chapter

Find three integrals in Exercises 27–70 for which a good strategy is to apply integration by parts twice.

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

∫u2du

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:∫(x-ex)2dx

Explain why ∫2xx2+1dxand ∫1xlnxdxare essentially the same integral after a change of variables.

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