Post on 31-Dec-2015
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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 substitutesin² x + cos² x = 1
sinm x • cosn x, n & m are even
Use half angle identitiessin² 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 IDcot2 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 IDsec² θ – 1 = tan² θ
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
Remove one sin x and combine with dx to form duUse Trig id: sin² x = 1 - cos² x to get the uⁿ du form
= ∫ sin² x (sin x dx)
= ∫ (1 – cos² x) (sin x dx)
= ∫ sin x dx – ∫ cos² x (sin x dx) )
= ∫ sin x dx – (- 1) ∫ u² du
= - cos x + ⅓ cos³ x + C
Let u = cos x then du = -sin x
7-2 Example 2
∫ cos5 xdx
let u = sin x and du = cos x dx
mostly un du form
= ∫ cos4 x (cos x dx)
= ∫ (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
= ∫ (1- sin2 x)2 (cos x dx)
= ∫ (1- 2sin2 x + sin4 x) (cos x dx)
Remove one cos x and combine with dx to form duUse Trig id: sin² x = 1 - cos² x to get the uⁿ du form
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
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 (1 - cos 2x) dx
= ½ x - ½(½ sin 2x) + C
= (¼) (2x – sin 2x) + C
= ½ ∫ dx - ½ ∫ cos 2x dx
7-2 Example 4
∫ cos4 x dx = ∫ cos4 x dx = ∫ (½(1 + cos 2x))²
= 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
Use double angle formulas:
cos2 x = ½(1 + cos 2x)
Twice on last term!
Then use cos u du forms
Note: Calculators will use other trig IDs to simplify into a different form
= ¼ (sin x cos³ x) + (3/8) sin x cos x + 3/8(x) + C
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 cos4 x dx =
let u = cos x and du = -sin x dx
∫ (1 – cos² x) (cos4 x) (sin x) dx =
= -1 ∫ (cos4 x) (-sin x) dx – (- ∫(cos6 x) (-sin x) dx)
= (-1/5) u5 + (1/7) u7 + C
= (-1/5) (cos5 x) + (1/7) (cos7 x) + C
= - ∫ u4 du + ∫ u6 du
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
Use ½ angle formulas
= ∫ (1/2) (1 – cos 2x) (1/2) (1 + cos 2x) dx
Have to use ½ angle formula again
= (1/4) ∫ (1 – cos2 2x) 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!)
= (1/4) ∫ dx – (1/4)∫ (1/2)(1 - cos 4x) dx
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 rule;other times, this may have to be repeated several times
7-2 Example 7
∫ cot4 x dx
Use trig id to convert cot2
= ∫ cot2 x (csc2 x – 1) dx
First is a u-sub power ruleand second, we reapply step 1
= ∫ cot2 x (csc2 x) dx – ∫ cot2 x dx
= - ∫ u² du - ∫ (csc2 x – 1) dx
= (-1/3)(cot3 x) + cot x + x + C
7-2 Example 8
Use trig id to convert cot2
= ∫ tan3 x (sec2 x – 1) dx
First is a u-sub power ruleand second, we reapply step 1
= ∫ tan3 x (sec2 x) dx – ∫ tan3 x dx
= ∫ 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
∫ tan5 x dx
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 andconvert other using trig id
= ∫ (tan-3/2 x) (tan2 + 1) (sec2 x) dx
= ∫ (tan1/2 x + tan-3/2 ) (sec2 x) dx
= ∫ 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
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 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
Using reduction formulas
= -(1/2) sin x cos x + (1/2) ∫ dx =
= (-1/2) sin x cos x + (1/2) x + C
Use your calculator to check.
Calculator uses the reduction formulas.
7-2 Example 11
Use trig reduction formula
= (1/5-1)tan5-1 x + ∫ tan5-2 x dx
Use trig reduction formula again
= (1/4) tan4 x + ∫ tan3 x dx
∫ tan5 x dx
= (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