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Trigonometric integrals: the whole story
A trigonometric integral is a (sum of) products or quotients of trig functions. Converting
tan x, sec x, etc. to sin’s and cos’s in the usual way, these integrals then come in 4 flavors:
Z
Z
Z
Z
sinn x
cosm x
1
n
m
sin x cos x dx,
dx,
dx,
dx
n
n
m
cos x
sin x
sin x cosm x
For each of these forms, there are (usually at least two!) generally successful methods for
“reducing” the integral to something which our other approaches can handle.
The basic underlying ideas are:
(1) Use u-substitution when possible to convert the problem to an integral of a polynomial
or rational (= quotient of polynomials) function.
(2) Use the trig identities
sin2 x + cos2 x = 1 , tan2 x + 1 = sec2 x , 1 + cot2 x = csc2 x (*)
1
1
1
and sin2 x = (1 − cos(2x)) , cos2 x = (1 + cos(2x)) , sin x cos x = sin(2x) (**)
2
2
2
to convert the integrand in order to help us reach an integral on which u-substitution can
be successful employed.
Z
More precisely: for
sinn x cosm x dx,
if n or m is odd, then
Z
Z
Z
m−1
n
n
m
m−1
sin x cos x dx =
sin x cos
x (cos x dx) =
sinn x (cos2 x) 2 (cos x dx) =
Z
m−1
=
sinn x (1 − sin2 x) 2 (cos x dx), or
Z
Z
Z
n−1
n
n−1
m
m
sin x cos x dx =
sin
x cos x (sin x dx) = (sin2 x) 2 cosm x (sin x dx) =
Z
n−1
− (1 − cos2 x) 2 cosm x (− sin x dx)
and the appropriate u-substitution will give us a polynomial to integrate.
If both n and m are even, then using the formulas (**) will yield trig integrals (in the
“variable” 2x) whose exponents are smaller; repeated use of these formulas will yield
integrals that the first technique can be applied to:
Z
Z
Z
1
1
2n
2
2m
n
2
m
sin x cos x dx = (sin x) (cos x) dx = ( (1−cos(2x)))n ( (1+cos(2x)))m dx
2
2
and multiply out! Odd exponents can be handled the first way; with even exponents we
can repeat this process again, obtaining trig integrals in the “variable ” 4x, etc.
Or! we can use cos2 x =Z1 − sin2 x to convert all of the cosines to sines.
Z
sin2n x cos2m x dx = sin2n x (1 − sin2 x)m x dx and multiply out. Then we can deal
with each term using a reduction formula:
Z
Z
n−1
−1
n−1
n
sinn−2 x dx (reduction formula)
sin
x cos x +
sin x dx =
n
n
1
a formula which can be discovered by using integration by parts:
Z
Z
n
sin x dx = sinn−1 x sin x dx , and set u = sinn−1 x and dv = sin x dx . We get
Z
n−1
− sin
x cos x + (n − 1) sinn−2 x cos2 x
Z
Z
n−1
n−2
= − sin
x cos x + (n − 1) sin
x dx − (n − 1) sinn x dx ;
we add the second term to the “other” side to get the formula above!
Z
sinn x
For
dx:
if n ≤ m then we can pair each sin x with a cos x to convert this to
cosm x
Z
Z
n
m−n
tan x sec
x dx =
tann x seck x dx.
[If n > m then we can replace one sin2 x on top with 1 − cos2 x to split the integral into
two integrals, one with two fewer sin x’s, and the other with two fewer of both sin x and
cos x. Repeated use of this will result in integrals of the forms we want: n ≤ m (together
with possibly an integral of the first type we looked at).]
These integrals can beZdealt with in a similar fashion. If
Z
Z k is even (and k ≥ 2), then
k−2
tann x seck x dx =
tann x seck−2 x (sec2 x dx) =
tann x(sec2 x) 2 (sec2 x dx) =
Z
k−2
tann x(tan2 x + 1) 2 (sec2 x dx)
and the substitution u = tan x will produce an integral of a polynomial to solve.
If
Z n is odd, then
Z
Z
n−1
n
k
n−1
k−1
tan x sec x dx = tan
x sec
x (sec x tan x dx) = (tan2 x) 2 seck−1 x (sec x tan x dx)
Z
n−1
= (sec2 x − 1) 2 seck−1 x (sec x tan x dx)
and the substitution u = sec x will produce a polynomial to integrate.
If
Z
Z k is odd and n is even,
Z then
n
n
k
n
k
2
tan x sec x dx = (tan x) 2 sec x dx = (sec2 −1) 2 seck x dx
which can be multiplied out to produce a sum of integrals of (odd) powers of sec x. Then
an integration by parts (integrating sec2 x, differentiating the rest) together with some
algebraic manipulation can be used to construct the reduction formula
Z
Z
n−2
1
n−2
n
secn−2 x dx
sec
x tan x +
sec x dx =
n−1
n−1
This allows us to continually reduce the exponent in our integral, until we reach
Z
sec2 x dx = tan x + C
Z
Z
Z
Z
sec x + tan x
sec2 x + sec x tan x
sec x tan x + sec2 x
sec x dx = sec x
dx =
dx =
dx
sec x + tan x
sec x + tan x
sec x + tan x
= ln | sec x + tan x| + C
2
Finally, if k = 0 [and n is even] (i.e., we have an integral of a power of tan x), then using
tan2 x = sec2 x−1 we can convert this to powers of sec x, and apply the previous approach.
Z
cosm x
For
dx:
sinn x
We could develop a completely parallel theory, using cot x and csc x instead of tan x and
π
π
sec x. Or we can use the substitution u = − x [so x = − u] and the fact that
2
Z 2 m
Z
π
cos x
sinm u
π
to rewrite
dx
=
(−du)
sin( −u) = cos u and cos( −u) = sin u
π ,
n
n
2
2
sin x
cos u
u= 2 −x
and apply the previous collection of techniques!
Z
1
For
dx,
we basically just do the best we can to end up somwhere else!
n
sin x cosm x
If
Z n or m is odd, then we
Z employ u-substitution: Z
sin x
sin x
1
dx =
dx
dx =
n+1
n
n+1
m
m
2
sin x cos x
sin
x cos x
(1 − cos x) 2 x cosm x
Z
Z
Z
1
cos x
cos x
or
dx
=
dx
=
m+1 dx
n
n
sin x cosm x
sin x cosm+1 x
(sinn x)(1 − sin2 x) 2
and the appropriate u-substitution will result in the integral of a rational function of u,
Z
du
(which we will learn later how to integrate!).
m+1 un (1 − u2 ) 2 u=sin x
1
sin(2x),
2
if n = m then we can rewrite the integral as an integral of a power of csc(2x), which
we can apply previous techniques to solve. [How? Just pretend that m − n = 0 in the
computations below.] If n > m or n < m, then we make the exponents in the denomenator
equal at the expense of putting a term in the numerator, and then convert:
m−n
Z
Z
m−n
Z 1
Z
( 2 (1 − cos(2x))) 2
sinm−n x
(sin2 x) 2
1
dx
dx =
dx =
dx =
sinn x cosm x
sinm x cosm x
(sin x cos x)m
( 21 sin(2x))m
n−m
Z
Z
Z
Z 1
n−m
( 2 (1 + cos(2x))) 2
1
cosn−m x
(cos2 x) 2
or
dx
dx =
dx =
dx =
sinn x cosm x
sinn x cosn x
(sin x cos x)n
( 12 sin(2x))n
If n and m are both even, then using the identity sin x cos x =
which can be treated (after multiplying out! and substitution u = 2x) as integrals of cos u’s
over sin u’s.
3
Summary of formulas:
Z
Z
m−1
n
m
sin x cos x dx =
sinn x (1 − sin2 x) 2 (cos x dx)
if m odd
Z
Z
n−1
n
m
if n odd
sin x cos x dx = − (1 − cos2 x) 2 cosm x (− sin x dx)
Z
Z
1
1
2n
2m
sin x cos x dx = ( (1 − cos(2x)))n ( (1 + cos(2x)))m dx
2
2
Z
Z
−1
n−1
sinn x dx =
sinn−2 x dx (reduction formula)
sinn−1 x cos x +
n
n
Z
Z
k−2
n
k
tan x sec x dx = (tann x)(tan2 x + 1) 2 (sec2 x dx)
if k ≥ 2 is even
Z
Z
n−1
n
k
tan x sec x dx = (sec2 x − 1) 2 seck−1 x (sec x tan x dx)
if n is odd
Z
Z
n
n
k
if k odd, n even
tan x sec x dx = (sec2 −1) 2 seck x dx
Z
Z
1
n−2
n
n−2
sec x dx =
secn−2 x dx
(reduction formula)
sec
x tan x +
n−1
n−1
Z
sec x dx = ln | sec x + tan x| + C
sinm u
du
π
cosn u
u= 2 −x
Z
cosm x
dx = −
sinn x
Z
1
dx =
n
sin x cosm x
Z
Z
sin x
dx
if n odd
n+1
(1 − cos2 x) 2 x cosm x
Z
Z
1
cos x
if m odd
dx =
m+1 dx
n
m
n
sin x cos x
(sin x)(1 − sin2 x) 2
m−n
Z
Z 1
[ 2 (1 − cos(2x))] 2
1
dx
if m ≥ n
dx =
sinn x cosm x
[ 21 sin(2x)]m
n−m
Z 1
Z
[ 2 (1 + cos(2x))] 2
1
dx =
dx
if n ≥ m
sinn x cosm x
[ 12 sin(2x)]n
Z
1
sin(mx) cos(nx) dx =
sin(m + n)x + sin(m − n)x dx
2
Z
Z
1
sin(mx) sin(nx) dx =
− cos(m + n)x + cos(m − n)x dx
2
Z
Z
1
cos(m + n)x + cos(m − n)x dx
cos(mx) cos(nx) dx =
2
Z
4