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Chapter 5: Q. 51i (page 465)

Solve the integral

鈭�(1-x2)-32dx

Short Answer

Expert verified

x-x2+1+C

Step by step solution

01

Step 1. Given

鈭�(1-x2)-32dxx-x2+1+C

02

Step 2. Using identity

To solve the integral x=sin(u)

now, derivation of the above equation is solved below

x=sin(u)

dx=cos(u)du

Now,usetheidentitysinh2(u)+1=cosh2(u).鈭�(1-x2)-32dx=鈭�1-sin2(u)+132=鈭�1cos2(u)32=鈭�1cos3(u)

03

Step 3. Calculation

鈭�(1-x2)-32dx=鈭�1cos2(u)du=鈭�sec2(u)du=tanu+C=tan(sin-1(x))+C=x-x2+1+C

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Most popular questions from this chapter

Consider the integral 鈭�sinxcosxdx.

(a) Solve this integral by using u-substitution with u=sinx and du=cosxdx.

(b) Solve the integral another way, using u-substitution with u=cosx and du=鈭�sinxdx.

(c) How must your two answers be related? Use algebra to prove this relationship.

Solve given definite integral.

鈭�鈭�11鈥�x39x2鈭�1dx

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

u(x)=x2+1

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 鈭� 1 at the bottom as the remainder. Thenp(x)=___andp(x)q(x)=____

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

x2鈭�4x鈭�8
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