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Unformatted text preview: 6-13SECTION 6.3..Trigonometric Techniques of Integration521what we mean, suppose thatf(x) andg(x) are functions withf(0)=g(0)=,f(1)=g(1)=0 and with continuous sec-ond derivativesf(x) andg(x). Use integration by parts twiceto show that1f(x)g(x)dx=1f(x)g(x)dx.3.Assume thatfis an increasing continuous function on [a,b]with 0≤a<bandf(x)≥0. LetA1be the area undery=f(x)fromx=atox=bandletA2betheareatotheleftofy=f(x)fromf(a)tof(b).ShowthatA1+A2=bf(b)−af(a)andbaf(x)dx=bf(b)−af(a)−f(b)f(a)f−1(y)dy.Usethisresult to evaluateπ/4tan−1x dx.yxaf(b)f(a)A2A1b4.Assume thatfis a function with a continuoussecond derivative. Show thatf(b)=f(a)+f(a)(b−a)+baf(x)(b−x)dx. Use thisresult to show that the error in the approximation sinx≈xis at most12x2.6.3TRIGONOMETRIC TECHNIQUES OF INTEGRATIONIntegrals Involving Powers of Trigonometric FunctionsEvaluating an integral whose integrand contains powers of one or more trigonometric func-tions often involves making a clever substitution. These integrals are sufficiently commonthat we present them here as a group.We first consider integrals of the formsinmxcosnx dx,wheremandnare positive integers.Case 1:mornIs an Odd Positive IntegerIfmis odd, first isolate one factor of sinx. (You’ll need this fordu.) Then, replace anyfactors of sin2xwith 1−cos2xand make the substitutionu=cosx. Likewise, ifnis odd,first isolate one factor of cosx. (You’ll need this fordu.) Then, replace any factors of cos2xwith 1−sin2xand make the substitutionu=sinx.We illustrate this for the case wheremis odd in example 3.1.EXAMPLE 3.1A Typical SubstitutionEvaluatecos4xsinx dx.SolutionSince you cannot evaluate this integral as it stands, you should considersubstitution. (Hint: Look for terms that are derivatives of other terms.) Here, lettingu=cosx, so thatdu= −sinx dx, gives uscos4xsinx dx= −cos4xu4(−sinx)dxdu= −u4du= −u55+c= −cos5x5+c.Sinceu=cosx.522CHAPTER 6..Integration Techniques6-14While this first example was not particularly challenging, it should give you an idea ofwhat to do with example 3.2.EXAMPLE 3.2An Integrand with an Odd Power of SineEvaluatecos4xsin3x dx.SolutionIf you’re looking for terms that are derivatives of other terms, you shouldsee both sine and cosine terms, but for which do you substitute? Here, withu=cosx,we havedu= −sinx dx, so thatcos4xsin3x dx=cos4xsin2xsinx dx= −cos4xsin2x(−sinx)dx= −cos4x(1−cos2x)u4(1−u2)(−sinx)dxdu= −u4(1−u2)du= −(u4−u6)du= −u55−u77+c= −cos5x5+cos7x7+c.Sinceu=cosx.Pay close attention to how we did this. We took the odd power (in this case, sin3x) andfactored out one power of sinx(to use fordu). The remaining (even) powers of sinxwere rewritten in terms of cosxusing the Pythagorean identitysin2x+cos2x=1....
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This note was uploaded on 10/03/2010 for the course CHE 2C CHE 2C taught by Professor Atsumi during the Spring '10 term at UC Davis.
- Spring '10