MTH 252 Integral CalculusOdd Powers of SIN & COSSlide 3Slide 4Slide 5Powers of SIN & COSPowers of TAN & SECEven Powers of SIN & COSSlide 9Powers of SECSlide 11Slide 12Powers of TANSlide 14FYI: More Reduction FormulasOne more note …MTH 252Integral CalculusChapter 8 – Principles of Integral EvaluationSection 8.3 – Trigonometric IntegralsCopyright © 2005 by Ron Wallace, all rights reserved.Odd Powers of SIN & COSsin nx dx�1sin sin nx x dx-=�n > 2 is a positive odd integern - 1 is a positive even integer( )1221 cos sin nx x dx-= -�( )122sin sin nx x dx-=�cossin u xdu x dx==-( )1221nu du-=- -�Multiply out the polynomial, integrate, and substitute back.Odd Powers of SIN & COScos nx dx�1cos cos nx x dx-=�n > 2 is a positive odd integern - 1 is a positive even integer( )1221 sin cos nx x dx-= -�( )122cos cos nx x dx-=�sincos u xdu x dx==( )1221nu du-= -�Multiply out the polynomial, integrate, and substitute back.Odd Powers of SIN & COSsin cos m nn x dx�1sin cos cos m nx x x dx-=�m is a positive integern is a positive odd integern - 1 is a positive even integer( )122sin 1 sin cos nmx x x dx-= -�( )122sin cos cos nmx x x dx-=�sincos u xdu x dx==( )1221nmu u du-= -�Multiply out the polynomial, integrate, and substitute back.Odd Powers of SIN & COSsin cos m nn x dx�1sin cos sin m nx x x dx-=�m is a positive odd integern is a positive integerm - 1 is a positive even integer( )1221 cos cos sin mnx x x dx-= -�( )122sin cos sin mnx x x dx-=�cossin u xdu x dx==-( )1221mnu u du-=- -�Multiply out the polynomial, integrate, and substitute back.Powers of SIN & COSsin cos m nn x dx�NOTE: m & n are non-negative integers.[ ] [ ]Since: sin cos & cos sind dx x x xdx dx= =-1. If n is odd, put one of the cosines w/ dx, change the remaining cosines to sines, and let u = sin x.2. If m is odd, put one of the sines w/ dx, change the remaining sines to cosines, and let u = cos x.3. All other cases … use some other method.2 21 sin cosx x- =2 21 cos sinx x- =Powers of TAN & SECtan sec m nn x dx�For what values of m & n can the same approach be used?[NOTE: m & n are non-negative integers.][ ] [ ]2Hint: tan sec & sec sec tand dx x x x xdx dx= =1. If m is odd and n > 0, put one of the tangents and one of the secants w/ dx, change the remaining tangents to secants, and let u = sec x.2. If n is even and n > 0, put two of the secants w/ dx, change the remaining secants to tangents, and let u = tan x.3. All other cases … use some other method.2 21 tan secx x+ =2 2sec 1 tanx x- =Even Powers of SIN & COSsin cos m nn x dx�m and n are BOTH non-negative even integers.Remember the half-angle formulas:1 coscos2 2q q+=�1 cossin2 2q q-=�1 cos 2cos2xx+=�1 cos2sin2xx-=�Let 2xq =21 cos2cos2xx+=21 cos 2sin2xx-=Square both sides.1.Use the trigonometric identities …2.Multiply everything out.3.Integrate each term, one at a time.Even Powers of SIN & COSsin cos m nn x dx�m and n are BOTH non-negative even integers.( )21cos 1 cos22x x= +( )21sin 1 cos 22x x= -Memorize this one!Powers of SECsec x dx�sec tansec sec tanx xx dxx x+� �=� �+� ��2sec sec tan sec tanx x xdxx x+=+�Let sec tanu x x= +2sec tan secdu x x x= +1 lndu u cu= = +�ln sec tanx x c= + +Powers of SECsec nx dx�2 2sec sec nu x dv x dx-= =sec ln sec tanx dx x x c= + +�n > 2 and a positive integer2sec tanx dx x c= +�Use Integration by Parts( )22 sec tan tanndu n x x dx v x-= - =2 2 2sec sec tan ( 2) sec tan n n nx dx x x n x x dx- -= - -� �( )2 2 2sec sec tan ( 2) sec sec 1 n n nx dx x x n x x dx- -= - - -� �2 2sec sec tan ( 2) sec ( 2) sec n n n nx dx x x n x dx n x dx- -= - - + -� � �2 2( 1) sec sec tan ( 2) sec n n nn x dx x x n x dx- -- = + -� �Powers of SECsec nx dx�2 2sec sec nu x dv x dx-= =sec ln sec tanx dx x x c= + +�n > 2 and a positive integer2sec tanx dx x c= +�Use Integration by Parts( )22 sec tan tanndu n x x dx v x-= - =2 2( 1) sec sec tan ( 2) sec n n nn x dx x x n x dx- -- = + -� �2 21 2sec sec tan sec 1 1n n nnx dx x x x dxn n- --= +- -� �This kind of identity is called a “Reduction Formula.”No! You do NOT need to memorize this one. Just use it.Memorize this one too!Powers of TANtan x dx�sin cosxdxx� �=� �� ��Let cosu x=sin du x dx=-1 lndu u cu=- =- +�ln sec x c= +ln cos x c=- +Powers of TANtan ln secx dx x c= +�tan nx dx�n > 1 and a positive integer1. If n is even, change to secants using …2. If n is odd, a. Put sec x tan x w/ dx as follows …b. Convert all tangents (except the one w/ dx) to secants.c. Use the substitution, u = sec x2 2tan sec 1x x= -1tantan sec tan secnnxx dx x x dxx-=� �FYI: More Reduction Formulas2 21 2sec sec tan sec 1 1n n nnx dx x x x dxn n- --= +- -� �1 21tan tan tan 1n n nx dx x x dxn- -= --� �1 21 1sin sin cos sin n n nnx dx x x x dxn n- --= +� �1 21 1cos cos sin cos n n nnx dx x x x dxn n- --= +� �One more note …1 1 sinConsider: ln2 1 sind xdx x+� �� �� �� �-� �� �( ) ( ) ( )( )21 sin cos 1 sin cos1 1 sin2 1 sin1 sinx x x xxxx� �- - + --� �=� �� �� �+� �-� �( )1 1 2cos2 1 sin 1 sinxx x� �� �=� �� �� �+ -� �� �( )2cos1 sinxx=-sec x=1 1 sinTherefore: sec ln2 1 sinxx dx cx+� �= +� �-� ��An equivalent form to the other