Review of Trigonometric Integrals

Review of Trigonometric Integrals - Review Trigonometric...

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Unformatted text preview: Review Trigonometric Integrals I. ∫cosnx dx Case 1: n is an odd natural number 1. ∫ cos x dx = sin x + c 2. cos 3 x dx ∫ = ∫ cos 2 x cos x dx = ∫ (1 − sin 2 x) cos x dx = ∫ (cos x − sin 2 x cos x) dx sin 3 x = sin x − +c 3 1* . sin 20 x cos x dx ∫ sin 21 x = +c 21 2 *. ∫ sin 20 3 x cos x dx = ∫ sin 20 x cos 2 x cos x dx = ∫ sin 20 x (1 − sin 2 x) cos x dx = ∫ (sin 20 x − sin 22 x ) cos x dx sin 21 x sin 23 x = − +c 21 23 cos 3 x ∫ 2 * *. 3 = ∫ sin − 2 3 = ∫ sin − 2 3 = ∫ (sin 1 3 − dx sin 2 x x cos 2 x cos x dx x (1 − sin 2 x) cos x dx 2 3 4 3 x − sin x ) cos x dx 7 3 sin x sin x = − +c 1 7 3 3 3. ∫ cos 5 x dx = ∫ cos x cos x dx 4 2 = ∫ (1 − sin x) cos x dx 2 = ∫ (1 − 2 sin x + sin x ) cos x dx 2 4 3 5 sin x sin x = sin x − 2 ⋅ + +c 3 5 4. ∫ cos 7 x dx = ∫ cos x cos x dx 6 3 = ∫ (1 − sin x) cos x dx 2 = ∫ (1 − 3 sin x + 3 sin x − sin x ) cos x dx 2 4 3 6 5 7 sin x 3 sin x sin x = sin x − 3 ⋅ + − +c 3 5 7 5 7 3 sin x sin x 3 = sin x − sin x + − +c 5 7 I. ∫cos x dx n Case 2: n is an even natural number 1. ∫ cos 2 x dx 1 + cos 2 x =∫ dx 2 1 = ∫ dx + ∫ cos 2 x dx 2 1 11 = x + ⋅ ∫ cos 2 x 2dx 2 22 1 1 = x + ⋅ sin 2 x + c 2 4 [ 2. ∫ cos 4 x dx 2 1 + cos 2 x I = ∫ dx 2 1 = 1 + 2 cos 2 x + cos 2 2 x dx ∫ 4 1 2 = ∫ dx + ∫ cos 2 x 2dx + ∫ cos 2 x dx 4 1 1 1 2 = x+ sin 2 x + ∫ cos 2 x dx 4 4 4 [ [ I 1 = ∫ cos 2 2 x dx 1 + cos 4 x dx 2 1 = ∫ dx + ∫ cos 4 x dx 2 1 11 = x + ⋅ ∫ cos 4 x 4dx 2 24 1 11 = x + ⋅ ⋅ sin 4 x + c 2 24 Thus, =∫ [ 1 1 11 1 I = x+ sin 2 x + x + sin 4 x + c 4 4 42 8 3 1 1 = x+ sin 2 x + sin 4 x + c 8 4 32 3. cos 2 x sin 2 x dx ∫ 1 + cos 2 x 1 − cos 2 x = ∫ dx 2 2 1 = ∫ 1 − cos 2 2 x dx 4 1 1 1 + cos 4 x = x− ∫ dx 4 4 2 1 1 1 = x − x + ∫ cos 4 x 4dx 2 8 4 3 1 = x − sin 4 x + c 8 32 ( ) II. ∫ tannx dx Case 1: n is an odd natural number 1. ∫ tan x dx sin x =∫ dx cos x − sin x = −∫ dx cos x = − ln cos x + c 2. tan 3 x dx ∫ = ∫ tan x tan x dx 2 = ∫ tan x (sec x − 1) dx 2 = ∫ (tan x sec 2 x − tan x) dx 1 = tan 2 x + ln cos x + c 2 3. tan 5 x dx ∫ = ∫ tan 3 x tan 2 x dx = ∫ tan 3 x (sec 2 x − 1) dx = ∫ (tan 3 x sec 2 x − tan 3 x) dx = ∫ tan 3 x sec 2 x dx − ∫ tan 3 x dx 1 1 4 = tan x − tan 2 x − ln cos x + c 4 2 ∫secmx tannx dx where n is odd 1. sec x tan x dx ∫ = sec x + c 2. sec x tan x dx ∫ 20 = ∫ sec x sec x tan x dx 19 20 sec x = +c 20 3. ∫ sec 20 3 x tan x dx = ∫ sec x (sec x − 1) sec x tan x dx 19 2 = ∫ (sec x − sec x) sec x tan x dx 21 22 19 20 sec x sec x = − +c 22 20 4. ∫ 2 3 5 sec x tan x dx 2 3 = ∫ sec x (sec 2 x − 1) 2 tan x dx = ∫ sec = ∫ sec −1 3 −1 3 x (sec 2 x − 1) 2 tan x sec x dx x (sec 4 x − 2 sec 2 x + 1) tan x sec x dx 11 3 5 3 = ∫ (sec x − 2 sec x + sec 14 3 8 3 2 3 −1 3 x) tan x sec x dx sec x sec x sec x = −2 + +c 11 8 2 3 3 3 II. ∫tan x dx n Case 2: n is an even natural number 1. tan x dx ∫ 2 = ∫ (sec x − 1 ) dx 2 = tan x − x + c 2. tan 4 x dx ∫ = ∫ tan 2 x (sec 2 x − 1 ) dx = ∫ (tan 2 x sec 2 x − tan 2 x ) dx tan 3 x = − ∫ tan 2 x d x 3 tan 3 x = − tan x + x + c 3 3. ∫ tan 6 x dx = ∫ tan x (sec x − 1 ) dx 4 2 = ∫ (tan 4 x sec 2 x − tan 4 x ) dx 5 tan x = − ∫ tan 4 x d x 5 5 3 tan x tan x = − + tan x − x + c 5 3 ∫secnx dx where n is even 1. sec x dx ∫ 2 = tan x + c 2. sec x dx ∫ 4 = ∫ (1 + tan x) sec x dx 2 2 = ∫ sec x dx + ∫ tan x sec x dx 2 2 3 tan x = tan x + +c 3 2 3. sec x dx ∫ 6 = ∫ (1 + tan x) sec x dx 2 2 2 = ∫ (1 + 2 tan x + tan x) sec x dx 2 4 3 2 5 tan x tan x = tan x + 2 ⋅ + +c 3 5 4. ∫ sec 8 x dx = ∫ (1 + tan x) sec x dx 2 3 2 = ∫ (1 + 3 tan x + 3 tan x + tan x) sec x dx 2 4 3 6 5 2 7 tan x tan x tan x = tan x + 3 ⋅ + 3⋅ + +c 3 5 7 5 7 tan x tan x 3 = tan x + tan x + 3 ⋅ + +c 5 7 ∫tanmx secnx dx where n is even 1. ∫ 3 tan 5 x sec 4 x dx 5 3 = ∫ tan x (1 + tan 2 x) sec 2 x dx 5 3 = ∫ (tan x + tan 8 3 14 3 11 3 x) sec 2 x dx tan x tan x = + +c 8 14 3 3 ∫secnx where n is odd 1. ∫ sec x dx sec x(tan x + sec x) =∫ dx tan x + sec x 2 sec x tan x + sec x =∫ dx tan x + sec x = ln tan x + sec x + c 2. 3. ∫ sec sec ∫ 3 x dx 5 x dx etc.. We use int egration by parts ...
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This note was uploaded on 10/12/2011 for the course MATH 201 taught by Professor Foad during the Spring '11 term at Qatar University.

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