Lesson 7-2
Hard Trig Integrals
Strategies for Hard Trig Integrals
Expression
Substitution
Trig Identity
sinn x or cosn x,
n is odd
Keep one sin x or cos x or for du
Convert remainder using Trig ID
sin² x + cos² x = 1
sinn x or cosn x,
n is even
Use half angle formulas:
sin² x = ½(1 – cos 2x)
cos² x = ½(1 + cos 2x)
sinm x • cosn x,
n or m is odd
From odd power,
keep one sin x or cos x, for du
Use identities to substitute
sin² x + cos² x = 1
sinm x • cosn x,
n & m are even
Use half angle identities
sin² x = ½(1 – cos 2x)
cos² x = ½(1 + cos 2x)
From power pull out tan2 x or cot2 x
and substitute using Trig ID
cot2 x = csc2 x - 1 or
tan2 x = sec2 x – 1
Pull out sec2 x or csc2 x for du
Convert rest to
tan or cot using the Trig ID
sec² θ – 1 = tan² θ
tann
x or
cotn
x
tanm x• secn x or
cotm x • cscn x ,
where n is even
Type I: sinn x or cosn x, n is odd
• Keep one sin x or cos x or for du
• Convert remainder with sin² x + cos² x = 1
• Using U substitution to get power rules
7-2 Example 1
∫ sin³ x dx
=
∫ sin² x (sin x dx)
Remove one sin x
and combine with dx to form du
Use Trig id: sin² x = 1 - cos² x
to get the uⁿ du form
=
∫ (1 – cos² x) (sin x dx)
=
∫ sin x dx – ∫ cos² x (sin x dx) )
=
∫ sin x dx – (- 1) ∫ u²
Let u = cos x then du = -sin x
= - cos x + ⅓ cos³ x + C
du
7-2 Example 2
∫ cos
5
xdx
Remove one cos x
and combine with dx to form du
Use Trig id: sin² x = 1 - cos² x
to get the uⁿ du form
let u = sin x and du = cos x dx
mostly un du form
∫
= (cos x dx) - 2
∫ cos
=
4
x (cos x dx)
∫
= (1- sin2 x)2 (cos x dx)
∫
= (1- 2sin2 x + sin4 x) (cos x dx)
∫ sin
2
x (cos x dx) +
= sin x – 2/3 sin3 x + 1/5 sin5 x + C
∫ sin
4
x (cos x dx)
Type 2: sinn x or cosn x, n is even
• Use half angle formulas:
– sin² x = ½(1 - cos 2x)
– cos² x = ½(1 + cos 2x)
• Use form of cos u du
7-2 Example 3
∫ sin² x dx
=
∫ ½ (1 - cos 2x) dx
Use double angle formulas:
Sin2 x = ½(1 – cos 2x)
Then use u = 2x and du = 2dx,
so you need an extra ½ out front
=½
∫ dx - ½ ∫ cos 2x dx
= ½ x - ½(½ sin 2x) + C
= (¼) (2x – sin 2x) + C
7-2 Example 4
∫ cos
4
x dx
=
∫ cos
4
x dx =
∫ (½(1 + cos 2x))²
Use double angle formulas:
cos2 x = ½(1 + cos 2x)
Twice on last term!
Then use
cos u du forms
∫
= ¼ (1 + 2cos 2x + cos2 2x) dx
= ¼( ∫ dx + 2 ∫ cos 2x dx + ∫ cos2 2x dx)
= ¼( ∫ dx + 2 ∫ cos 2x dx + ∫ ½(1 + cos 4x) dx)
= ¼x + ¼sin 2x + (1/8)x + (1/8)(1/4) sin 4x + C
= (3/8)x + ¼sin 2x + (1/32) sin 4x + C
= ¼ (sin x cos³ x) + (3/8) sin x cos x + 3/8(x) + C
Note: Calculators will use other trig IDs to simplify into a different form
Type 3: sinm x • cosn x, n or m is odd
• From odd power,
keep one sin x or cos x, for du
• Use identities to substitute
– Convert remainder with sin² x + cos² x = 1
• With U-substitutions, use power rule
7-2 Example 5
∫ sin³ x cos
4
let u = cos x and
du = -sin x dx
∫ (1 – cos² x) (cos
x dx =
4
= -1
=-
∫ (cos
∫u
4
4
x) (sin x) dx =
∫
x) (-sin x) dx – (- (cos6 x) (-sin x) dx)
∫
du + u6 du
= (-1/5) u5 + (1/7) u7 + C
= (-1/5) (cos5 x) + (1/7) (cos7 x) + C
Type IV: sinm x • cosn x, n and m are even.
• Use half angle identities
– sin² x = ½(1 - cos 2x)
– cos² x = ½(1 + cos 2x)
7-2 Example 7
∫ sin² x cos² x dx
∫
= (1/2) (1 – cos 2x) (1/2) (1 + cos 2x) dx
Use ½ angle formulas
∫
= (1/4) (1 – cos2 2x) dx
Have to use ½ angle formula again
∫
∫
= (1/4) dx – (1/4) (1/2)(1 - cos 4x) dx
= (1/4) x – (1/8) x + (1/8)
∫ cos 4x dx
= (1/8) x + (1/32) sin 4x + C
(not similar to calculator answer!)
Type V: tann x or cotn x
• From power pull out tan2 x or cot2 x
and substitute cot2 x = csc2 x - 1
or tan2 x = sec2 x – 1
• Sometimes it converts directly into usubstitution and the power rule;
other times, this may have to be repeated
several times
7-2 Example 7
∫ cot
4
x dx
∫
= cot2 x (csc2 x – 1) dx
Use trig id to convert cot2
∫
∫
= cot2 x (csc2 x) dx – cot2 x dx
First  is a u-sub power rule
and second, we reapply step 1
=-
∫ u² du - ∫ (csc
2
x – 1) dx
= (-1/3)(cot3 x) + cot x + x + C
7-2 Example 8
∫ tan
5
x dx
∫
= tan3 x (sec2 x – 1) dx
Use trig id to convert cot2
∫
∫
= tan3 x (sec2 x) dx – tan3 x dx
First  is a u-sub power rule
and second, we reapply step 1
=
∫u
du - tan x(sec2 x – 1) dx
=
∫u
du - u du + tan x dx
3
3
∫
∫
∫
= (1/4)(tan4 x) - (1/2)tan2 x - ln |cos x| + C
Type VI: tanm x• secn x or cotm x • cscn x ,
where n is even
• Pull out sec2 x or csc2 x for du
• Convert rest using trig ids:
– csc2 x = cot2 x + 1
– sec2 x = tan2 x + 1
• Use u-substitution and power rules
7-2 Example 9
∫ tan
-3/2
x sec4 x dx
Keep a sec2 for du and
convert other using trig id
∫
= (tan-3/2 x) (tan2 + 1) (sec2 x) dx
∫
= (tan1/2 x + tan-3/2 ) (sec2 x) dx
=
∫u
1/2
du +
∫u
-3/2
du
= (2/3)u3/2 – (2) u-1/2 + C
= (2/3)tan3/2 x – (2)tan-1/2 x + C
Trigonometric Reduction Formulas
Expression
∫ sinn x dx
∫ cosn x dx
∫ tann x dx
∫ secn x dx
Reduction Formula
1
n–1
n-1
= - --- sin x cos x + ------- ∫ sinn-2 x dx
n
n
1
n–1
n-1
= --- cos x sin x + ------- ∫ cosn-2 x dx
n
n
1
= ------- tann-1 x - ∫ tann-2 x dx
n-1
1
n-2
n-2
= ------- sec x tan x + ------- ∫ secn-2 x dx
n-1
n-1
Remember the following integrals: (when n=1 in the above)
∫ tan x dx = ln |sec x| + C
∫ sec x dx = ln |sec x + tan x| + C
7-2 Example 10
∫ sin² x dx
∫
= -(1/2) sin x cos x + (1/2) dx =
Using reduction formulas
= (-1/2) sin x cos x + (1/2) x + C
Use your calculator to check.
Calculator uses the reduction formulas.
7-2 Example 11
∫ tan
5
x dx
∫
= (1/5-1)tan5-1 x + tan5-2 x dx
Use trig reduction formula
∫
= (1/4) tan4 x + tan3 x dx
Use trig reduction formula again
∫
= (1/4) tan4 x + (1/3-1)tan3-1 x + tan3-2 x dx
∫
= (1/4) tan4 x + (1/2)tan2 x + tan x dx
= (1/4) tan4 x + (1/2)tan2 x + ln|sec x| + C
Summary & Homework
• Summary:
– Hard Trig integrals can be solved
• Homework:
– pg 488-489,
Day 1: 1, 2, 5, 9, 10
Day 2: 3, 7, 11, 14, 17