Section 8.1 Basic Integration Rules Fitting Integrands to Basic Rules

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332460_0801.qxd 11/2/04 3:04 PM Page 519SECTION 8.1 Basic Integration Rules 519EXAMPLE 2Evaluate10Using Two Basic Rules to Solve a Single Integralx 34 x 2 dx.x + 3 2yy =4 − x 21Solution Begin by writing the integral as the sum of two integrals. Then apply thePower Rule and the Arcsine Rule as follows.111x 304 x dx x2 0 4 x dx 3 1 1114 x22 0 4 x dx 22 12 2x dx 3 2 2 x dx 200−1The area of the region is approximately1.839.Figure 8.11xSee Figure 8.1. 4 x 2 12 3 arcsin x 2 1 0 3 1.8392 2 0TECHNOLOGY Simpson’s Rule can be used to give a good approximation ofthe value of the integral in Example 2 (for the approximation is 1.839).When using numerical integration, however, you should be aware that Simpson’sRule does not always give good approximations when one or both of the limits ofintegration are near a vertical asymptote. For instance, using the FundamentalTheorem of Calculus, you can obtain1.990n 10,x 34 x dx 6.213. 2Applying Simpson’s Rule (with n 10) to this integral produces an approximationof 6.889.EXAMPLE 3 A Substitution Involving a 2 u 2xFind 216 x dx. 6STUDY TIP Rules 18, 19, and 20 of thebasic integration rules on the next pageall have expressions involving the sum ordifference of two squares:a 2 u 2a 2 u 2u 2 a 2With such an expression, consider thesubstitution as in Example 3.u f x,SolutionBecause the radical in the denominator can be written in the forma 2 u 2 4 2 x 3 2you can try the substitution u x 3 . Then du 3x 2 dx, and you havex2Rewrite integral.16 x dx 1 3x32 dx6 16 x 1 duSubstitution: u x33 24 32 u 2 1 Arcsine Rule3 arcsin u 4 C 1 3arcsinx34 C.Rewrite as a function of x.
332460_0801.qxd 11/2/04 3:04 PM Page 520520 CHAPTER 8 Integration Techniques, L’Hôpital’s Rule, and Improper IntegralsSurprisingly, two of the most commonly overlooked integration rules are the LogRule and the Power Rule. Notice in the next two examples how these two integrationrules can be disguised.Review of Basic IntegrationRules a > 01. kfu du kf u du2. f u ± gu du 3. du u C4.5.f u du ±gu duu n du un1n 1 C, du u ln u Cn 1EXAMPLE 4FindA Disguised Form of the Log RuleSolution The integral does not appear to fit any of the basic rules. However, thequotient form suggests the Log Rule. If you let u 1 e x , then du e x dx. You canobtain the required du by adding and subtracting e x in the numerator, as follows.11 e x dx.11 e x dx 1 ex e x1 e x dx 1 e x1 e x e x dx e x dx1 e x x ln1 e x CAdd and subtract e x in numerator.x Rewrite as two fractions.1 e dxRewrite as two integrals.Integrate.6.7.8.9.10.e u du e u Ca u du 1ln a au Csin u du cos u Ccos u du sin u Ctan u du ln cos u CNOTE There is usually more than one way to solve an integration problem. For instance, inExample 4, try integrating by multiplying the numerator and denominator by e x to obtain anintegral of the form duu. See if you can get the same answer by this procedure. (Becareful: the answer will appear in a different form.)EXAMPLE 5 A Disguised Form of the Power RuleFind cot xlnsin x dx.11. cot u du ln sin u C12.13.14. sec 2 u du tan u C15. csc 2 u du cot u C16. sec u tan u du sec u C17.18.19.sec u du ln sec u tan u Ccsc u du ln csc u cot u Ccsc u cot u du csc u Cdua 2 u arcsin u 2 a Cdua 2 u 1 2 a arctan u a C20. duuu 2 a 2 1 a arcsec u a CSolution Again, the integral does not appear to fit any of the basic rules. However,considering the two primary choices for u cot x and u lnsin x, you can seethat the second choice is the appropriate one becauseSo,NOTEuu lnsin xand cot xlnsin x dx u duSubstitution: u lnsin x u2Integrate.2 CIn Example 5, try checking that the derivative of12 lnsin x2 Cis the integrand of the original integral.du cos xsin x dx 1 2 lnsin x2 C. cot x dx.Rewrite as a function of x.
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Section 8.1 Basic Integration Rules Fitting Integrands to Basic Rules

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