Integrating powers of products of sec x and tan x
Andrés Eduardo Caicedo
February 1, 2011
(In what follows, I will write tanZn x for (tan x)n , secm x for (sec x)m , etc.)
In lecture we saw how to find
sinn (x) cosm (x) dx for n, m = 0, 1, . . . I
want to explain here how a similar approach allows us to find
Z
secm x · tann x dx,
for m, n = 0, 1, . . .
This is a good example of the usefulness of reduction formulas.
Recall that
1
sin x
and sec x =
,
tan x =
cos x
cos x
that
(tan x)0 = sec2 x and (sec x)0 = sec x · tan x,
and that
sec2 x = tan2 x + 1.
The formulas below make use of these identities repeatedly.
1
Integrating powers of sec x
Recall that
Z
sec x dx = ln | sec x + tan x| + C
and
Z
sec2 x dx = tan x + C.
Let me explain how to obtain the following reduction formula for integrating
powers of sec x:
Z
Z
secm−2 x · tan x m − 2
secm x dx =
+
secm−2 x dx + C,
m−1
m−1
this formula is valid for any m ≥ 2.
1
The idea is to use integration by parts. Suppose m ≥ 2. Then
Z
Z
secm x dx = secm−2 x sec2 x dx.
Letting f (x) = secm−2 x and g 0 (x) = sec2 (x) in (∗):
Z
Z
f g 0 = f g − f 0 g,
we have f g 0 = secm x, f g = secm−2 x tan x, and
f 0g
= (m − 2) secm−3 x sec x tan x · tan x = (m − 2) secm−2 x tan2 x
= (m − 2) secm−2 x(sec2 x − 1) = (m − 2) secm x − (m − 2) secm−2 x.
Replacing in (∗), we have
Z
Z
Z
secm x dx = secm−2 x tan x − (m − 2) secm x dx + (m − 2) secm−2 x dx,
or
Z
(m − 1)
secm x dx = secm−2 x tan x + (m − 2)
Z
secm x dx + C,
and dividing both sides by m − 1 gives us the reduction formula
Z
Z
secm−2 x · tan x m − 2
secm x dx =
+
secm−2 x dx + D.
m−1
m−1
Z
Z
4
Exercise 1. Find
sec x dx and
sec5 x dx.
2
Integrating powers of tan x
We already know that
Z
tan x dx = ln | sec x| + C.
The corresponding reduction formula is
Z
Z
tann−1 x
− tann−2 x dx,
tann x dx =
n−1
valid for all n ≥ 2.
To see this, note that tann x = tann−2 x tan2 x = tann−2 x(sec2 x − 1) =
n−2
tan
x sec2 x − tann−2 x, so
Z
Z
Z
n
n−2
2
tan x dx = tan
x sec x dx − tann−2 x dx,
and the first integral on the right hand side can be easily computed using the
substitution u = tan x.
Z
Z
4
Exercise 2. Find
tan x dx and
tan5 x dx.
2
3
Integrating products of powers of sec x and
tan x
Suppose now that we need to evaluate an expression of the form
Z
secm x · tann x dx,
where both n and m are at least 1. As in the case of integrals of products of
powers of sines and cosines, it is best to divide the problem into two cases.
3.1
If n is odd
Suppose first that n is odd, say n = 2k + 1, for some integer k ≥ 0. Then
Z
Z
secm x · tann x dx = secm−1 x · tan2k x · (sec x · tan x) dx.
Since tan2 x = sec2 x − 1, we can write tan2k x = (sec2 x − 1)k . The last integral
can then be expressed in the form
Z
secm−1 x · (sec2 x − 1)k · (sec x · tan x) dx.
This can be easily evaluated using the substitution u = sec x, that transforms
it into
Z
um−1 (u2 − 1)k du.
To evaluate this expression, expand (u2 − 1)k and multiply the result by um−1 .
This gives us a polynomial in u. We can integrate the polynomial term by term,
and then replace sec x back in place of u.
Z
Z
Exercise 3. Find
sec3 x · tan x dx and
sec4 x · tan5 x dx.
3.2
If n is even
Suppose now that n is even, say n = 2k for some integer k ≥ 1. Then
Z
Z
secm x · tann x dx = secm x (sec2 x − 1)k dx.
To evaluate this expression, expand (sec2 x − 1)k and multiply the result by
secm x. This gives us a sum of powers of sec x, that can be evaluated term by
term using the reduction formula from Section 1. Note that this method works
even if m = 0.
Z
Exercise 4. Find
sec3 x · tan4 dx.
Z
Exercise 5. Find
tan4 x dx using this method, and show that your answer
actually gives the same result as the answer you found in Exercise 2.
3
4
Integrating powers of csc x and cot x
Exercise 6. (This is long.) Explain how to
Z adapt the methods from the previous
sections to find any integral of the form
cscm x · cotn x dx. Then repeat the
previous exercises but with csc in place of sec, and cot instead of tan .
For some extra credit, you have until Friday, February 11 at the beginning
of lecture, to turn in as many of the exercises above as you want.
4