Trigonometric integrals
Previous lecture
Z
1
sin x cos x =
2
sin2 x dx =
Z
1
2
1
cos x dx =
2
1
sin 2x dx = cos 2x + C,
4
(1 cos 2x) dx =
x 1
sin 2x + C
2 4
(1 + cos 2x) dx =
x 1
+ sin 2x + C
2 4
Integrals of the form
Z
sin ax cos bx dx,
sin ax sin bx dx,
cos ax cos bx dx,
a 6= b
Two basic identities:
sin (x + y) = sin x cos y + cos x sin y
cos (x + y) = cos x cos y sin x sin y
(for small x, y sin is increasing, +, cos is decreasing, )
Change y to y
sin (x y) = sin x cos y cos x sin y
cos (x y) = cos x cos y + sin x sin y
2 sin x cos y = sin (x + y) + sin (x y)
2 cos x cos y = cos (x + y) + cos (x y)
2 sin x sin y = cos (x y) cos (x + y)
Application to integrals.
Example
Z
1
sin 2x cos 3x dx =
2
=
1
2
(sin (2x + 3x) + sin (2x 3x)) dx =
MATH115 T2 Winter 04, A.Potapov
sin 5x dx
1
2
sin x dx =
1
1
cos 5x + cos x + C.
10
2
1
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Example
Z
cos 3x cos 5x dx =
=
1
2
1
2
(cos (3x + 5x) + cos (3x 5x)) dx =
cos 8x dx +
1
2
cos 2x dx =
1
1
sin 8x + sin 2x + C.
16
4
=
Example
Z
1
sin 3x sin 5x dx =
2
1
=
2
(cos (3x 5x) cos (3x + 5x)) dx =
1
cos 2x dx
2
cos 8x dx =
1
1
sin 2x
sin 8x + C.
4
16
Integrals of the form
Z
sinm x cosn x dx,
tanm x secn x dx,
cotm x cscn x dx.
Idea use substitution to transform to integral of polynomial
Z
Pk (u)du
or
Pk (u)
ds.
us
Actual substitution depends on m, n, and the type of the integral.
Typical substitutions: u = sin x, cos x, tan x, sec x, cot x, csc x.
Task: find, which substitution integral can be transformed
Even or odd powers in
P (u)du without
1 u2 or similar.
cosm x sinn x dx
cosm x sinn x dx = cosm1 x sinn x cos xdx
cos xdx = d (sin x), u = sin x, but have to express cosm1 x through sin x.
If m = 2k + 1 (odd) we can: cos2k x = cos2 x
If m = 2k (even) we can not without
= 1 sin2 x
= 1 u2
1 sin2 x.
Similarly
cosm x sinn x dx = cosm x sinn1 x sin xdx
MATH115 T2 Winter 04, A.Potapov
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If n = 2k + 1 (odd) we can use u = cos x, if n = 2k (even) cannot.
sincos rule: if one has odd power, use other for substitution.
1) Odd power at cos x, u = sin x, du = cos x dx, cos2 x = 1 u2
Z
cos2k+1 x sinn x dx =
cos2k x sinn x cos xdx =
1 u2
un du
Example: no sin x, but still u = sin x
Z
cos x
cos x dx =
Z
cos x dx =
1 u2
du =
2
1
1 2u2 + u4 du = u u3 + u5 + C =
3
5
= sin x
2
1
sin3 x + sin5 x + C.
3
5
2) Odd power at sin x, u = cos x, du = sin x dx, sin2 x = 1 u2
Z
Z
m
2k+1
cos x sin
Z
m
x dx =
2k
cos x sin
um 1 u2
x sin xdx =
du
3) Odd powers at both: use for u one with bigger power:
Example: a) u = cos x
Z
cos3 x sin9 x dx =
Z
cos3 x sin8 x sin xdx =
u3 1 u2
du =
u3 1 4u2 + 6u4 4u6 + u8 du =
4
6
4
1
1
cos10 x
cos12 x + C
= cos4 x + cos6 x cos8 x +
4
6
8
10
12
b) u = sin x
Z
Z
3
cos x sin x dx =
Z
u9 u11 du =
cos x sin x cos xdx =
1 u2 u9 du =
1 10
1
1
1
u u12 + C =
sin10 x
sin12 x + C
10
12
10
12
which is simpler?
4) Even powers at both. No good substitution. Transform with double angle formulas (x 2x) to smaller
powers:
sin2 x =
1 cos 2x
,
2
MATH115 T2 Winter 04, A.Potapov
cos2 x =
1 + cos 2x
,
2
sin x cos x =
sin 2x
2
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Example:
Z
sin4 x cos6 xdx =
Z
=
1
=
64
(sin x cos x)4 cos2 x =
1
1 + cos 2x 1
sin4 2x
d (2x) = [z = 2x]
16
2
2
1
sin zdz [double angle] +
64
4
1
=
64
1
=
256
1 cos 2z
2
sin4 z cos zdz [u = sin z] =
1
dz +
64
u4 du =
1 2 cos 2z + cos2 2z dz +
Z
1 5
u +C =
320
z
1
1
sin z +
256 256
512
x
1
x
1
1
sin 2x +
+
sin 4z +
sin5 2x + C =
128 256
256 2048
320
=
1
sin5 z + C =
320
3
1
1
1
x
sin 2x +
sin 8x +
sin5 2x + C.
256
256
2048
320
R
Even or odd powers in
(1 + cos 4z) dz +
tanm x secn x dx
If m = 2k + 1 is odd, you can rewrite and use the substitution u = cos x
Z
Z
2k+1
tan
x sec x dx =
sin2k+1 x
dx =
cos2k+1+n x
1 u2
du.
u2k+1+n
Example:
Z
tan x dx =
sin x
dx [u = cos x] =
cos x
du
= ln |cos x| + C
u
Example:
Z
Z
3
tan x dx =
Z
Z
3
du +
sin3 x
dx = [u = cos x] =
cos3 x
1 u2
du =
u3
1
1
u1 du = u2 + ln |u| + C = sec2 x + ln |cos x| + C
2
2
Other possibility to use substitutions
a)
u = tan x,
du = sec2 x dx,
sec2 x = 1 + u2
so the power at sec x must be even, power at tan x may be any;
b)
u = sec x,
du = tan x sec x dx,
tan2 x = u2 1
So the power at tan x must be odd, power at sec x may be any;
MATH115 T2 Winter 04, A.Potapov
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1) Even power at sec x, n = 2k + 2, u = tan x,
Z
tanm x sec2k+2 x dx =
um 1 + u2
du.
Example:
Z
tan4 x sec4 x dx =
u4 1 + u2 du =
u4 + u6 du =
1
1
1
1
= u5 + u7 + C = tan5 x + tan7 x + C
5
7
5
7
Example: no tan x, but still u = tan x
Z
Z
4
sec x dx =
sec x sec x dx =
1 + u2 du =
1
1
= u + u3 + C = tan x + tan3 x + C.
3
3
2) Odd power at tan x, m = 2k + 1, u = sec x,
Z
tan2k+1 x secn x dx =
tan2 x
secn1 x tan x sec xdx =
u2 1
un1 du.
Example:
Z
tan1 x sec0 xdx =
tan xdx =
(sec x)1 tan x sec xdx =
u1 du =
= ln |u| + C = ln |sec x| + C = ln |cos x| + C
Other types:
tan2k+2 x dx
Reduction formula
Z
tan
Z
tan2k x sin2 x sec2 x dx =
x dx =
tan2k x 1 cos2 x sec2 x dx =
=
Z
2k+2
Z
2k
tan
x dx =
MATH115 T2 Winter 04, A.Potapov
Z
2k
u du +
tan2k x sec2 x dx [u = tan x] +
tan2k2 x 1 cos2 x sec2 x dx = ...
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Other types:
seck x dx, k = 1, 3, 5, ...
R
Denote for brevity Jk =
seck x dx
k=1
Z
J1 =
sec xdx =
Z
dx
=
cos x
du
1
=
2
1u
2
1
1
+
du =
1+u 1u
cos xdx
= [u = sin x]
cos2 x
1
1 1 + u
1 1 + sin x
(ln |1 + u| ln |1 u|) + C = ln
+ C = ln
+C =
2
2
1u
2
1 sin x
(1 + sin x)
(1 + sin x)2
1 (1 + sin x)2
1
+C =
+ C = ln
= ln
+ C = ln
2
2
(1 sin x) (1 + sin x)
2
cos x
1 sin x
= ln |sec x + tan x| + C
k 3: reduction formula
Z
Jk =
seck x dx =
seck2 x sec2 xdx =
seck2 x (tan x)0 dx =
= sec
k2
x tan x
tan x (k 2) seck3 x tan x sec xdx =
Z
= seck2 x tan x (k 2)
= seck2 x tan x (k 2)
tan2 x seck2 xdx =
sec2 x 1 seck2 xdx =
= seck2 x tan x (k 2) (Jk Jk2 )
Reduction formula
Jk =
1
k2
seck2 x tan x +
Jk2
k1
k1
Application:
J3 =
Integrals of Ik =
1
1
1
1
sec x tan x + J1 = sec x tan x + ln |sec x + tan x| + C
2
2
2
2
csck x dx can be evaluated in a similar way..
MATH115 T2 Winter 04, A.Potapov
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