Hard Trig Integrals

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Lesson 7-2

• Hard Trig Integrals

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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)
tann x or cotn x 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
tanm x secn x or cotm x cscn x , where n is even Pull out sec2 x or csc2 x for duConvert rest to tan or cot using the Trig ID sec² ? 1 tan² ?
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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

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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 du - cos x ? cos³ x C
Let u cos x then du -sin x
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7-2 Example 2
? cos5 xdx
? cos4 x (cos x dx)
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
? (1- sin2 x)2 (cos x dx)
? (1- 2sin2 x sin4 x) (cos x dx)
let u sin x and du cos x dx mostly un
du form
? (cos x dx) - 2 ? sin2 x (cos x dx) ? sin4 x
(cos x dx) sin x 2/3 sin3 x 1/5 sin5 x C
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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

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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
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7-2 Example 4
? cos4 x dx ? (½(1 cos 2x))²
? cos4 x dx
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
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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

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7-2 Example 5
? sin³ x cos4 x dx
? (1 cos² x) (cos4 x) (sin x) dx
-1 ? (cos4 x) (-sin x) dx (- ?(cos6 x) (-sin
x) dx)
let u cos x and du -sin x dx
- ? u4 du ? u6 du
(-1/5) u5 (1/7) u7 C (-1/5)
(cos5 x) (1/7) (cos7 x) C
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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)

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7-2 Example 7
? (1/2) (1 cos 2x) (1/2) (1 cos 2x) dx
? sin² x cos² x 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!)
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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
  u-substitution and the power ruleother times,
  this may have to be repeated several times

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7-2 Example 7
? cot2 x (csc2 x 1) dx
? cot4 x 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 - ? (csc2 x 1) dx (-1/3)(cot3
x) cot x x C
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7-2 Example 8
? tan3 x (sec2 x 1) dx
? tan5 x 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
? u3 du - ? tan x(sec2 x 1) dx ? u3 du
- ? u du ? tan x dx (1/4)(tan4 x) -
(1/2)tan2 x - ln cos x C
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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

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7-2 Example 9
? (tan-3/2 x) (tan2 1) (sec2 x) dx
? tan-3/2 x sec4 x dx
? (tan1/2 x tan-3/2 ) (sec2 x) dx
Keep a sec2 for du and convert other using trig id
? u1/2 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
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Trigonometric Reduction Formulas
Expression Reduction Formula
? sinn x dx 1 n 1 - --- sinn-1 x cos x ------- sinn-2 x dx n n
? cosn x dx 1 n 1 --- cosn-1 x sin x ------- cosn-2 x dx n n
? tann x dx 1 ------- tann-1 x - tann-2 x dx n - 1
? secn x dx 1 n - 2 ------- secn-2 x tan x ------- secn-2 x dx n - 1 n - 1
?
?
?
?
Remember the following integrals (when n1 in
the above) ? tan x dx ln sec x C
? sec x dx ln sec x tan x C
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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.
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7-2 Example 11
(1/5-1)tan5-1 x ? tan5-2 x dx
? tan5 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 lnsec x C
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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