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Chapter 5: Q. 39 (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�x1/4sinx5/4dx

Short Answer

Expert verified

The solution of the given integral is 鈭�x1/4sinx5/4dx=-45cosx5/4+C.

Step by step solution

01

Step 1. Given Information 

Solving the given integrals.

鈭�x1/4sinx5/4dx

02

Step 2. Solving the given integral using substitution method.  

Let

u=x5/4dudx=54x1/4du=54x1/4dx45du=x1/4dx

03

Step 3. This substitution changes the integral into  

鈭�x1/4sinx5/4dx=45鈭�sinudu鈭�x1/4sinx5/4dx=-45cosu+C鈭�x1/4sinx5/4dx=-45cosx5/4+C

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Most popular questions from this chapter

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

u(x)=1x

Solve given definite integral.

鈭�35鈥�x2鈭�9dx

Give an example of an integral for which trigonometric substitution is possible but an easier method is available. Then give an example of an integral that we still don鈥檛 know how to solve given the techniques we know at this point.

Consider the integral 鈭�1x21鈭�x2dxfrom the reading at the beginning of the section.

(a) Use the inverse trigonometric substitution u=sin鈭�1鈦�xto solve this integral.

(b) Use the trigonometric substitution x=sin鈦�u to solve the integral.

(c) Compare and contrast the two methods used in parts (a) and (b).

Why don鈥檛 we ever have cause to use the trigonometric substitution x=cosu?
See all solutions

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