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Title
Trigonometric integrals
MATH 1700
October 31, 2016
MATH 1700
Trigonometric integrals
October 31, 2016
1/1
Readings
Readings
Readings: Section 7.2
MATH 1700
Trigonometric integrals
October 31, 2016
2/1
R
One odd power in
R
sinm x cosn xdx
sinm x cosn x
• If the power of cos x is odd, use cos2 x = 1 − sin2 x to eliminate all
but one cos x:
Z
Z
m
2k +1
sin x cos
x = sinm x(1 − sin2 x)k cos xdx.
Substitute u = sin x.
MATH 1700
Trigonometric integrals
October 31, 2016
3/1
R
One odd power in
R
sinm x cosn xdx
sinm x cosn x
• If the power of cos x is odd, use cos2 x = 1 − sin2 x to eliminate all
but one cos x:
Z
Z
m
2k +1
sin x cos
x = sinm x(1 − sin2 x)k cos xdx.
Substitute u = sin x.
• If the power of sin x is odd, use sin2 x = 1 − cos2 x to eliminate all
but one power of sin x:
Z
Z
2k +1
n
sin
x cos xdx = sin x(1 − cos2 x)k cosn xdx.
Substitute u = cos x.
MATH 1700
Trigonometric integrals
October 31, 2016
3/1
R
sinm x cosn xdx
Both powers even in
R
sinm x cosn x
If both powers of sin x and cos x are both even, use the double-angle
formulas
sin2 x =
MATH 1700
1
1
(1 − cos 2x) or cos2 x = (1 + cos 2x) .
2
2
Trigonometric integrals
October 31, 2016
4/1
R
sinm x cosn xdx
Both powers even in
R
sinm x cosn x
If both powers of sin x and cos x are both even, use the double-angle
formulas
sin2 x =
1
1
(1 − cos 2x) or cos2 x = (1 + cos 2x) .
2
2
Sometimes this double-angle formula is also useful:
sin 2x = 2 sin x cos x.
MATH 1700
Trigonometric integrals
October 31, 2016
4/1
R
R
tanm x secn xdx
tanm x secn xdx
• If the power of sec x is even, use sec2 x = 1 + tan2 x to eliminate all
but sec2 x:
Z
Z
m
2k
tan x sec xdx = tanm x(1 + tan2 x)k −1 sec2 xdx.
Substitute u = tan x.
MATH 1700
Trigonometric integrals
October 31, 2016
5/1
R
R
tanm x secn xdx
tanm x secn xdx
• If the power of sec x is even, use sec2 x = 1 + tan2 x to eliminate all
but sec2 x:
Z
Z
m
2k
tan x sec xdx = tanm x(1 + tan2 x)k −1 sec2 xdx.
Substitute u = tan x.
• If the power of tan x is odd, use tan2 x = sec2 x − 1 to write
everything in terms of sec x except for one sec x tan x:
Z
Z
2k +1
n
tan
x sec xdx = (sec2 x − 1)k secn−1 x sec x tan x.
Substitute u = sec x.
MATH 1700
Trigonometric integrals
October 31, 2016
5/1
R
R
tanm x secn xdx
tanm x secn xdx
In other cases, these tricks sometimes help:
Z
tan xdx = ln | sec x| + C
Z
sec xdx = ln | sec x + tan x| + C.
MATH 1700
Trigonometric integrals
October 31, 2016
6/1
R
R
sin mx cos nx,
R
tanm x secn xdx
cos mx cos nx,
R
To evaluate any of the integrals
Z
Z
sin mx cos nx,
cos mx cos nx,
sin mx sin nx
Z
sin mx sin nx
use one of these identities:
1
(sin (A − B) + sin (A + B))
2
1
sin A sin B = (cos (A − B) − cos (A + B))
2
1
cos A cos B = (cos (A − B) + cos (A + B)) .
2
sin A cos B =
MATH 1700
Trigonometric integrals
October 31, 2016
7/1