INTEGRATION BY PARTS

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1 Mathmatics Rvision Guids Intgration by Parts Pag of 7 MK HOME TUITION Mathmatics Rvision Guids Lvl: AS / A Lvl AQA : C Edcl: C OCR: C OCR MEI: C INTEGRATION BY PARTS Vrsion : Dat: --5 Eampls - 6 ar copyrightd to thir rspctiv ownrs and usd with thir prmission

2 Mathmatics Rvision Guids Intgration by Parts Pag of 7 INTEGRATION BY PARTS This tchniqu is a rvrs rsult of th proct rul for diffrntiation Rcall: If y uv, whr u and v ar functions of, thn dy d uv) u v d d d d ( Intgrating with rspct to, w hav uv u d d v d d Rarranging, w hav th standard formula for intgration by parts, i u d d uv v d d ) g'( ) d f ( ) g( ), or, in function notation, f g( ) f '( ) d ( In ordr to b abl to us this mthod, th function d must b intgrabl Additionally, th proct v d should not b mor difficult to intgrat than th original Gnrally, if on of th factors of th proct is a polynomial, thn st u to b th polynomial, bcaus its dgr will b rcd by by ach application of th procss Eampl (): Find i) / 6 sin d ii) sin d, laving th answr as a surd in trms of i) Tak u d Tak d sin v -cos sin d cos ( cos ) d cos cos d cos sin c ii) Th dfinit intgral is workd out in th usual way: / 6 sin d cos sin / 6 ( cos ( )) sin ( )) ((cos ()) sin( )) ( 6 6 6

3 Mathmatics Rvision Guids Intgration by Parts Pag of 7 Eampl (): Find i) d ii) d, laving th rsult in trms of i) Lt u d Lt d d ( ) d d c ( ) c v ½ ( ) ii) Th dfinit intgral is ( ) Somtims th procss might nd to b carrid out mor than onc Eampl (): Find cos d Lt u d Lt d cos v sin cos d sin sin d Th intgrand on th right-hand sid nds anothr application of th procss, but not how th dgr of th polynomial (th u function) has bn rcd from (quadratic) to (linar) Lt u d Lt d sin v -cos sin d cos cos d cos sin c Th two rsults nd combining to giv th final intgral: (watch out for th + and signs!) cos d sin cos sin sin cos sin c

4 Mathmatics Rvision Guids Intgration by Parts Pag of 7 Eampl (): i) Show that d ii) Us th rsult from i) to find ar intgr constants (, givn that d ) c ) c d (, giving your answr in th form a b - whr a and b i) Two applications ar ndd hr First application: Tak u d Tak d - v - - d ( ) d Labl th right-hand intgrand I Scond application: This intgrand I is (-) tims th on quotd in part i), so th intgral of I ( ) c d ( ) c ( ) c ( ) c ii) To find d, w work out th indfinit intgral first : Tak u d Tak d - v - - d ( ) d Labl th right-hand intgrand I Th intgrand I on th RHS is (-) tims th intgrand ( ) c Th complt intgral d d, so its intgral is ( ) c ( ) c ( 6 6) c Hnc d ( 6 6) (8 6) - ( 6) 6 8 -

5 Mathmatics Rvision Guids Intgration by Parts Pag 5 of 7 Intgrals involving procts of ln nd to b handld a littl diffrntly, sinc ln is not asy to intgrat, but it can b diffrntiatd In thos cass, ln should b takn as u Eampl (5): Find ln d Tak u ln d Tak d v ln d ln d ln d ln c 6 Somtims an prssion which dos not look lik a proct can b dfind as on, by trating as a trm Eampl (6): Find ln d Trat th prssion as ln Thus u ln d Tak d v ln d (ln ) ( ) d ln d ln c

6 Mathmatics Rvision Guids Intgration by Parts Pag 6 of 7 Eampl (7): i) Evaluat 7 tan d ii) Th vlocity v of a prformanc car (in mtrs pr scond) is modlld by th quation v 8tan t whr radian masur is usd Us your knowldg of transformations to valuat distanc travlld in thos 6 sconds Giv your rsult in mtrs to significant figurs 6 8tan (t) dt, and hnc calculat th Trat th prssion as tan - Thus u tan - d Tak d v 7 7 tan d tan 7 tan ln( ) 7 tan 7 ln(58) d 7 ii) Th intgrand 6 8tan ( ) dt t is obtaind by transforming th intgrand 7 tan d by an -strtch of scal factor of th variabl from to t followd by a y-strtch of scal factor 8, plus a trivial rnaming Now 8 7, so w multiply th rsult of part i) by 7 to obtain th distanc travlld by th car in thos 6 sconds namly 8 7, or 58 mtrs to significant figurs

7 Mathmatics Rvision Guids Intgration by Parts Pag 7 of 7 Eampl (8): Find Hr u d sin d Also d sin v -cos sin d ( cos ) ( cos ) cos cos d Applying intgration by parts again w hav u d d cos v sin cos d (sin ) (sin ) sin sin d () Combining th two rsults, w finally hav sin d cos sin sin d Th procss appars to b going on for vr without simplifying th prssion, but notic how th original intgrand now appars on both sids of th quation Adding this intgrand to both sids will cancl it out from th RHS sin d cos sin sin d ( sin cos ) c () From th abov rsults in () and () w can also dc that cos d cos d sin ( sin cos ) ( sin cos ) c

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