IV. Summary
∫ sin
[ I.B ]
n
x dx ,
∫ cos
n
x dx
n even (n = 2p) -write
€
sin n x = sin 2 p x = (sin 2 x) p = (
1 − cos 2x p
)
2
= 1p (1 − cos 2x) p , or
2
cosn x = 1p (1 + cos 2x) p , expand algebraically and integrate term-by-term as
2
powers of cos 2x
€
n odd (n = 2p+1) -n
n−1
write sin x = sin x ⋅ sin x
= sin 2 p x ⋅ sin x = (1 − cos2 x) p ⋅ sin x ,
2 p
use the substitution u = cos x , du = −sin x dx → (1 − u ) ⋅ (−du) , and expand
€
€
algebraically;
n
2
or cos x = (1 − sin
x) p ⋅ cos x , u = sin x , du = cos x dx → (1 − u 2 ) p ⋅ du
any integer n > 1 -can also use the reduction formulae
€
€
€
1
n−1
n
n−1
n−2
sin
x
dx
=
−
sin
x
cos
x
+
sin
x dx ,
∫
n
n ∫
n−2
x dx
∫ cosn x dx = 1n cosn−1 x sin x + n−1
n ∫ cos
€
∫ sin
m
x cos n x dx
€
m even ( m = 2p )
m odd ( m = 2p + 1 )
€
n even ( n = 2q )
I.D
I.C.1
n odd ( n = 2q + 1 )
I.C.2
I.C.1 or I.C.2
sin m x cos n x = sin 2 p x cos n x ⋅ sin x = (1 − cos 2 x) p cos 2q x ⋅ sin x ,
2 p
2q
use the substitution u = cos x , du = −sin x dx → (1 − u ) ⋅ u ⋅ (−du) , and
I.C.1: write
expand algebraically
€ write
I.C.2:
sin m x cos n x = sin m x cos2q x ⋅ cos x = sin m x (1 − sin 2 x) q ⋅ cos x ,
€ x dx → u 2 p ⋅ (1 − u 2 ) q ⋅ du , and
use the substitution u = sin x , du = cos
expand algebraically
I.D:€write
sin m x cos n x = sin 2 p x cos 2q x , which can be expressed as either
€ 2 x) q ; expand algebraically and integrate
(1 − cos2 x) p ⋅ cos2q x or sin 2 p ⋅ (1 − sin
term-by-term as powers of sin2x or cos2x , using the results in I.B
m
m
2p
special case ( m = n = 2p ) -- can also write sin x cos x = (sin x cos x)
€
1 − cos 4 x p
= ( 12 sin2x) 2 p = 12 p (sin 2 2x) p = 12 p (
) = 12 p ⋅ 1p (1 − cos 4 x) p
2
2
2
2
2
= 13 p (1 − cos 4 x) p , expand algebraically and integrate term-by-term as
2
€
powers of cos 4x
€
€
€
∫ tan
n
n
∫ sec x dx
n
n
(results are analogous for ∫ cot x dx , ∫ csc x dx
[ II.B ]
€
x dx ,
secant (any integer n > 1) -n
n−2
write sec x = sec
x ⋅ sec 2
)
x and use integration by parts with u = secn−2 x ,
dv = sec2x dx ; the
€ integration produces a term containing the original integral;
can also be used to obtain the reduction formula
€
∫ sec
n
1
x dx = n−1
sec n−2 x tan x + n−2
n−1
tangent (any integer n > 1) -n
n−2
write tan x = tan
x ⋅ tan 2 x
∫ sec
n−2
x dx
= tan n−2 x ⋅ (sec 2 x − 1) and use integration by
−
parts€on one of the terms, with u = tann 2 x , dv = sec2x dx ; the integration
produces a term containing the original integral; can also be used to obtain the
reduction formula
€
∫ tan
n
1
x dx = n−1
tan n−1 x −
tangent for n even ( n = 2p ) -n
2p
can also write tan x = tan x
∫ tan
n−2
x dx
= (sec 2 x − 1) p , expand algebraically and
€ as powers of sec2 x , using the results above
integrate
∫ tan€
m
x sec n x dx
m even ( m = 2p )
m odd ( m = 2p + 1 )
€
n
n even ( n = 2q )
II.C
II.C or II.D
m
∫ cot
m
x csc n x dx
)
n odd ( n = 2q + 1 )
II.E
II.D
€
tan x sec x = tan x sec x = tan m x sec 2q−2 x ⋅ sec 2 x ,
m
2
q−1
use the substitution u = tan x , du = sec2x dx → u ⋅ ( u + 1) ⋅ du , and
II.C: write
m
(results are analogous for
2q
expand algebraically
€ write
II.D:
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 ,
2
2p
n−1
use the substitution u = sec x , du = €
sec x tan x dx → ( u −1) ⋅ u ⋅ du , and
expand algebraically
€ write tan x sec x = tan x sec x = (tan x) sec x = (sec x − 1) sec
II.E:
€
expand algebraically and integrate term-by-term
as powers of sec x , using the
results in II.B
m
€
n
2p
n
2
p
n
2
p
n
x ,
-- G. Ruffa
12-21 September 2003
revised and amended 25-26 January 2009