trigintegrals_001

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IV. Summary

[ I.B ] sin n x dx " , cos n x dx "

n even ( n = 2 p ) --

write sin n x = sin 2 p

x = (sin 2 x ) p

= (1 " cos 2 x

2 ) p=

12 p

(1 " cos 2 x ) p , or

cos n x =

12 p

(1 + cos 2 x ) p , expand algebraically and integrate term-by-term as

powers of cos 2 x

n odd ( n = 2 p +1) --write sin n

x = sin n " 1 x # sin x = sin 2 p

x # sin x = (1 " cos 2 x ) p # sin x ,

use the substitution u = cos x , du = ! sin x dx " (1 # u2 ) p $ (# du ) , and expandalgebraically;or cos n

x = (1 " sin 2 x ) p # cos x , u = sin x , du = cos x dx " (1 # u2 ) p $ du

any integer n > 1 --can also use the reduction formulae

sin n x dx = " 1n # sin n "1 x cos x +

n "1n sin n " 2 x dx # ,

cos n x dx =1n " cos n #1 x sin x +

n #1n cos n # 2 x dx "

sin m x cos n x dx "

n even ( n = 2 q ) n odd ( n = 2 q + 1 ) m even ( m = 2 p ) I.D I.C.2m odd ( m = 2 p + 1 ) I.C.1 I.C.1 or I.C.2

I.C.1: write sin m x cos n

x = sin 2 p x cos n

x " sin x = (1 # cos 2 x ) p cos 2 q

x " sin x ,

use the substitution u = cos x , du = ! sin x dx " (1 # u2 ) p $ u 2 q $ (# du) , andexpand algebraically

I.C.2: write sin m x cos n

x = sin m x cos 2 q

x " cos x = sin m x (1 # sin 2

x ) q " cos x ,

use the substitution u = sin x , du = cos x dx " u2 p # (1 $ u 2 ) q # du , andexpand algebraically

I.D: write sin m x cos n

x = sin 2 p x cos 2 q

x , which can be expressed as either

(1 " cos 2 x ) p # cos 2 q x or sin 2 p # (1 " sin 2 x ) q ; expand algebraically and integrate

term-by-term as powers of sin 2x or cos 2x , using the results in I.B special case ( m = n = 2 p ) -- can also write sin m

x cos m x = (sin x cos x ) 2 p

= ( 12 sin2 x ) 2 p

=1

2 2 p(sin 2 2 x ) p

=1

2 2 p(1 " cos 4 x

2 ) p=

12 2 p

# 12 p

(1 " cos 4 x ) p

=1

2 3 p(1 " cos 4 x ) p , expand algebraically and integrate term-by-term as

powers of cos 4 x



[ II.B ] tann x dx " , sec

n x dx "

(results are analogous for cotn x dx " , csc

n x dx " )

secant (any integer n > 1) --

write secn

x = secn " 2

x # sec2

x and use integration by parts with u = secn! 2

x ,dv = sec 2x dx ; the integration produces a term containing the original integral;can also be used to obtain the reduction formula

sec n x dx =1

n " 1 # sec n " 2 x tan x +n " 2n " 1

sec n " 2 x dx #

tangent (any integer n > 1) -- write tan n

x = tan n " 2 x # tan 2

x = tan n " 2 x # (sec 2

x " 1) and use integration byparts on one of the terms, with u = tan n! 2 x , dv = sec 2x dx ; the integrationproduces a term containing the original integral; can also be used to obtain thereduction formula

tan n x dx =

1n "1 # tan n " 1 x " tan n " 2 x dx #

tangent for n even ( n = 2 p ) --can also write tan n

x = tan 2 p x = (sec 2

x " 1) p , expand algebraically andintegrate as powers of sec 2 x , using the results above

tanm x sec

n x dx " (results are analogous for cotm x csc

n x dx " )

n even ( n = 2 q ) n odd ( n = 2 q + 1 ) m even ( m = 2 p ) II.C II.Em odd ( m = 2 p + 1 ) II.C or II.D II.D

II.C: write tan m x sec n

x = tan m x sec 2 q

x = tan m x sec 2 q " 2

x # sec 2 x ,

use the substitution u = tan x , du = sec 2x dx " um # ( u 2+ 1) q $ 1 # du , and

expand algebraically

II.D: write tan m x sec n

x = tan 2 p + 1 x sec n

x = tan 2 p x sec n " 1

x # sec x tan x ,

use the substitution u = sec x , du = sec x tan x dx " ( u 2 # 1) 2 p $ u n # 1 $ du , andexpand algebraically

II.E: write tan m x sec n

x = tan 2 p x sec n

x = (tan 2 x ) p sec n

x = (sec 2 x " 1) p sec n

x , expand algebraically and integrate term-by-term as powers of sec x , using theresults in II.B

-- G. Ruffa12-21 September 2003

revised and amended 25-26 January 2009

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