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 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

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