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UCR - MATH 009B - Math 9B, Week 5 - Class Notes

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UCR - MATH 009B - Math 9B, Week 5 - Class Notes

Description: These notes go over trigonometric integrals and the strategies for integrating products of sines and cosines as well as products of tangents and secants.
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Join more than 18,000+ college students at University of California Riverside who use StudySoup to get ahead
Description: These notes go over trigonometric integrals and the strategies for integrating products of sines and cosines as well as products of tangents and secants.
2 Pages 29 Views 23 Unlocks
• Notes, Study Guides, Flashcards + More!

Unformatted text preview: 6.3 Trigonometric Integrals We are given the integral in one of two forms _product of sines and cosines sin" (x) cosm @dx Product of tangents and secants stan sech ()dx We transform it into an easy one of the form Binu.cos(x) dx Scos"Q) sin @dx Stan sea tan@dx 2 We transform the given integral into an easy integral using the Fria identties: ? Sin? QP+cos? (x) = tan?(x) =sec (x) sin?(x)= 1-cos(2x) cos?Q=1tcos(2x) 2 Recalli: :sing)'cosc) 'tan 'esec() cos(x)=sin(x) sec(x)=secatan(x) Ex.l jsin^(.cos(x)dx use u-substitution with sin@) as u .its derivative is cos@ so it will go away = fundu - Lutte now that it's integrated, bang back the original variable = sinnt (x) +CEx. 2 Ssecni tandx chanele into an easy integral slundu set y to be seca so seca tan@ goes away it und te int- seen HD+C Strategies to integrate more complex products of sines and cosines 1. If n = 2k+ is odd, use sin" ()= |-C05 sin" (cos(x)=sin). (sin Wyk. Cosmo = sin() [l-cos (x)]k.cos" @ u-cos given duEsindx easy by substitution 2. If m=2K+i oddy use cos?) = | sin ) 3in" (cos(x)=sin([cos(x)]k.cos) a sin^@[l-sin? (6)]5. cos() uasin (. given du e-cos@dx easy bu substitution 3. If n = 2k and m=2P both even. use sin Q)-l-cos(2) and cos? () = 1cos(2x) Sin"@cosm()= [sin?(x)]K. [cos?_)]' = l 1-cos(2x) * ( cos(2x)) 2 given weEx3 Scos (2x) dx 3 is odd 30 we separate one cosine and transform all other cosines to sines using cos? (0) - I-sin? (0) = s cos(2x) cos?(2x) dx = 3 cos (2x) [i-sin?(2x)]dx apply u-substitution u=sin (2x) du-cos(lx) 2dx 42 du=cos (2x) dx = ] Il-? 1.72 du integrate as usual now that it's usingu - Uz S(1-3 du osplit yms. (I-updu into two integrals =Y2E5ldu- su du] 542 Sidu-Y2 Surdu =Y 2.-42:13u> rc replace u with the original variable =Y2 Sin (2x) - Yousins () +C Ex.4 Isip .cos@dx :3 is odd so we separate one sine, and turn all other = sin(x) sin?X.cos(xclx = Isina). [l-cos?)].cos? @dx. apply u-substitution du=-sin(x) dx -du= sindx. u=00SX ay u Brys us to --/3 cos@+y5.cos W C

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