Trigonometric Integrals - Bruce E. Shapiro

TOPIC 7. TRIGONOMETRIC INTEGRALS Math 150A
∫
Example 7.9
tan 5 x sec 7 x dx
∫
d(sec x)
∫
{ }} {
tan 5 x sec 7 x dx = tan 4 x sec 6 x (sec x tan x dx)
∫
= tan 4 x sec 6 x d(sec x)
∫
= (sec 2 x − 1) 2 sec 6 x d(sec x)
∫
= (sec 4 x − 2 sec 2 x + 1) sec 6 x d(sec x)
∫
= (sec 10 x − 2 sec 8 x + sec 6 x) ( .
sec x)
(7.23a)
(7.23b)
(7.23c)
(7.23d)
(7.23e)
= 1
11 sec11 x − 2 9 sec9 x + 1 7 sec7 x + C (7.23f)
Sometimes we can get somewhere by combining integration by parts with
an appropriate substitution, as in the following.
∫
Example 7.10 sec 3 x dx We start by using integration by parts, with
u = sec x =⇒ du = sec x tan x dx (7.24a)
dv = sec 2 x dx =⇒ v = tan x (7.24b)
With these substitutions
∫
∫
sec 3 x dx = uv − v du
∫
= sec x tan x − tan x sec x tan x dx
∫
= sec x tan x − tan 2 x sec x dx
∫
= sec x tan x − (sec 2 x − 1) sec x dx
∫
∫
= sec x tan x − sec 3 x dx + sec x dx
(7.24c)
(7.24d)
(7.24e)
(7.24f)
(7.24g)
Since the original integral appears on both sides of the equation, but with
opposite signs, we can add.
∫
∫
2 sec 3 x dx = sec x tan x + sec x dx
(7.24h)
= sec x tan x + ln | sec x + tan x| (7.24i)
« 2012. Last revised: February 26, 2013 Page 27