π Activity 5.3.1. π πConsider .β«sinβ‘(x)cosβ‘(x)dx. Which substitution would you choose to evaluate this integral? u=sinβ‘(x) u=cosβ‘(x) u=sinβ‘(x)cosβ‘(x) Substitution is not effective
π Activity 5.3.2. π πConsider .β«sin4β‘(x)cosβ‘(x)dx. Which substitution would you choose to evaluate this integral? u=sinβ‘(x) u=sin4β‘(x) u=cosβ‘(x) Substitution is not effective
π Activity 5.3.3. π πConsider .β«sin4β‘(x)cos3β‘(x)dx. Which substitution would you choose to evaluate this integral? u=sinβ‘(x) u=cos3β‘(x) u=cosβ‘(x) Substitution is not effective
π Activity 5.3.4. πItβs possible to use substitution to evaluate ,β«sin4β‘(x)cos3β‘(x)dx, by taking advantage of the trigonometric identity .sin2β‘(x)+cos2β‘(x)=1. πComplete the following substitution of u=sinβ‘(x),du=cosβ‘(x)dx by filling in the missing ?s. β«sin4β‘(x)cos3β‘(x)dx=β«sin4β‘(x)(?)cosβ‘(x)dx=β«sin4β‘(x)(1β?)cosβ‘(x)dx=β«?(1β?)du=β«(u4βu6)du=15u5β17u7+C=?
π Activity 5.3.5. πTrying to substitute u=cosβ‘(x),du=βsinβ‘(x)dx in the previous example is less successful. β«sin4β‘(x)cos3β‘(x)dx=ββ«sin3β‘(x)cos3β‘(x)(βsinβ‘(x)dx)=ββ«sin3β‘(x)u3du=β―? π πWhich feature of sin4β‘(x)cos3β‘(x) made u=sinβ‘(x) the better choice? The even power of sin4β‘(x) The odd power of cos3β‘(x)
π Activity 5.3.6. π πTry to show β«sin5β‘(x)cos2β‘(x)dx=β17cos7β‘(x)+25cos5β‘(x)β13cos3β‘(x)+C πby first trying ,u=sinβ‘(x), and then trying u=cosβ‘(x) instead. π πWhich substitution worked better and why? u=sinβ‘(x) due to sin5β‘(x)βs odd power. u=sinβ‘(x) due to cos2β‘(x)βs even power. u=cosβ‘(x) due to sin5β‘(x)βs odd power. u=cosβ‘(x) due to cos2β‘(x)βs even power.
π Observation 5.3.7. πWhen integrating the form :β«sinmβ‘(x)cosnβ‘(x)dx: If sinβs power is odd, rewrite the integral as β«g(cosβ‘(x))sinβ‘(x)dx and use .u=cosβ‘(x). If cosβs power is odd, rewrite the integral as β«h(sinβ‘(x))cosβ‘(x)dx and use .u=sinβ‘(x).
π Activity 5.3.8. πLetβs consider .β«sin2β‘(x)dx. π(a) πUse the fact that sin2β‘(ΞΈ)=1βcosβ‘(2ΞΈ)2 to rewrite the integrand using the above identities as an integral involving .cosβ‘(2x). π(b) πShow that the integral evaluates to .12xβ14sinβ‘(2x)+C.
π(a) πUse the fact that sin2β‘(ΞΈ)=1βcosβ‘(2ΞΈ)2 to rewrite the integrand using the above identities as an integral involving .cosβ‘(2x).
π Activity 5.3.9. πLetβs consider .β«sin2β‘(x)cos2β‘(x)dx. π(a) πUse the fact that cos2β‘(ΞΈ)=1+cosβ‘(2ΞΈ)2 and sin2β‘(ΞΈ)=1βcosβ‘(2ΞΈ)2 to rewrite the integrand using the above identities as an integral involving .cos2β‘(2x). π(b) πUse the above identities to rewrite this new integrand as one involving .cosβ‘(4x). π(c) πShow that integral evaluates to .18xβ132sinβ‘(4x)+C.
π(a) πUse the fact that cos2β‘(ΞΈ)=1+cosβ‘(2ΞΈ)2 and sin2β‘(ΞΈ)=1βcosβ‘(2ΞΈ)2 to rewrite the integrand using the above identities as an integral involving .cos2β‘(2x).
π Activity 5.3.10. π πConsider .β«sin4β‘(x)cos4β‘(x)dx. Which would be the most useful way to rewrite the integral? β«(1βcos2β‘(x))2cos4β‘(x)dx β«sin4β‘(x)(1βsin2β‘(x))2dx β«(1βcosβ‘(2x)2)2(1+cosβ‘(2x)2)2dx
π Activity 5.3.11. π πConsider .β«sin3β‘(x)cos5β‘(x)dx. Which would be the most useful way to rewrite the integral? β«(1βcos2β‘(x))cos5β‘(x)sinβ‘(x)dx β«sin3β‘(x)(1+cosβ‘(2x)2)2cosβ‘(x)dx β«sin3β‘(x)(1βsin2β‘(x))2cosβ‘(x)dx
π Remark 5.3.12. πWe might also use some other trigonometric identities to manipulate our integrands, listed in Appendix B.
π Activity 5.3.13. πConsider .β«sinβ‘(ΞΈ)sinβ‘(3ΞΈ)dΞΈ. π(a) πFind an identity from Appendix B which could be used to transform our integrand.π(b) πRewrite the integrand using the selected identity.π(c) πEvaluate the integral.