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MATH 137 Trigonometric Integrals 1. Standard Trig Functions: (i) ∫ sin x dx = − cos x + C (ii) ∫ cos x dx = sin x + C (iii) ∫ tan x dx = − ln cos x + C (iv) ∫ cot x dx = ln sin x + C Proof. Let u = cos x . Then and du = − sin x dx Proof. du = dx . Then − sin x sin x sin x du ∫ tan x dx = ∫ cos x dx = ∫ u × − sin x = −∫ 1 du = − ln u + C u = − ln cos x + C . (v) ∫ sec x dx = ln sec x + tan x + C (vi) ∫ csc x dx = − ln csc x + cot x + C Proof. We alter the integral as follows: Proof. sec x + tan x ∫ sec x dx = ∫ sec x × sec x + tan x dx =∫ sec 2 x + sec x tan x dx . sec x + tan x Now let u = sec x + tan x so that we du have = sec x tan x + sec 2 x . Then dx ∫ sec x dx = ∫ sec 2 x + sec x tan x dx sec x + tan x sec 2 x + sec x tan x du × u sec x tan x + sec 2 x 1 = ∫ du = ln u + C = ln sec x + tan x + C. u =∫ 2. Standard Hyperbolic Trig Functions: (i) ∫ sinh x dx = cosh x + C (ii) ∫ cosh x dx = sinh x + C (iii) ∫ tanh x dx = ln(cosh x) + C (iv) ∫ coth x dx = ln sinh x + C Proof. Let u = cosh x . Then and du = sinh x dx Proof. du = dx . Then sinh x sinh x sinh x du ∫ tanh x dx = ∫ cosh x dx = ∫ u × sinh x =∫ 1 du = ln u + C u = ln cosh x + C = ln(cosh x ) + C, because cosh x > 0 for all x . (v) ∫ sech x dx = 2 tan −1 (e x ) + C (vi) ∫ csch x dx = −2 tanh −1(e x ) + C Proof. We alter the integral as follows: Proof. 1 2 ∫ sech x dx = ∫ cosh x dx = ∫ x − x dx e +e =∫ 2 2 dx = ∫ 2x dx 1 e +1 ex + x e ex ex = 2 ∫ 2x dx e +1 Now let u = e x so that du = e x dx . Then ex sech x dx = 2 dx ∫ ∫ 2x e +1 1 = 2∫ 2 du = 2 tan −1 (u) + C u +1 = 2 tan −1 (e x ) + C 3. Squares of Standard Trig Functions: We need the following identities: cos2 x + sin 2 x = 1 1 + tan 2 x = sec 2 x cot 2 x + 1 = csc 2 x cos2 x − sin 2 x = cos(2x ) or (1 − sin 2 x ) − sin 2 x = cos(2x ) , which gives 2 1 − 2sin 2 x = cos(2 x) , and then sin x = 1 − cos(2 x) . 2 Likewise, cos2 x − (1 − cos2 x ) = cos(2x ) , which gives 2 2 cos2 x − 1 = cos(2 x) , and then cos x = x 1 − sin(2x ) + C 2 4 2 (i) ∫ sin x dx = 2 Proof. Using sin x = 2 1 + cos(2 x) . 2 (ii) ∫ cos x dx = x 1 + sin(2x ) + C 2 4 1 − cos(2 x) Proof. , we have 2 1 1 2 ∫ sin x dx = ∫ 2 − 2 cos(2x ) dx 1 1 1 x − × sin(2 x) + C 2 2 2 x 1 = − sin(2 x) + C . 2 4 = (iii) ∫ sec 2 x dx = tan x + C (iv) ∫ csc 2 x dx = − cot x + C (v) ∫ tan 2 x dx = tan x − x + C (vi) ∫ cot 2 x dx = − cot x − x + C 2 2 Proof. Using 1 + tan x = sec x , we have 2 2 ∫ tan x dx = ∫ (sec x − 1) dx = tan x − x + C. Proof. 4. Squares of Standard Hyperbolic Trig Functions: We need the following identities: cosh 2 x − sinh 2 x = 1 1 − tanh 2 x = sech 2 x coth 2 x − 1 = csch 2 x cosh 2 x + sinh2 x = cosh(2x ) or (1 + sinh 2 x ) + sinh 2 x = cosh(2 x) , which gives 2 1 + 2 sinh2 x = cosh(2x ) , and then sinh x = cosh(2 x) − 1 . 2 Likewise, cosh 2 x + (cosh 2 x − 1) = cosh(2x ) , which gives 2 2 cosh 2 x − 1 = cosh(2x ) , and then cosh x = 2 (i) ∫ sinh x dx = Proof. With 1 x sinh(2 x) − + C 4 2 sinh2 x = have cosh(2 x) − 1 , 2 1 2 ∫ sinh x dx = ∫ 2 cosh(2 x ) − 2 cosh(2 x) + 1 . 2 (ii) ∫ cosh x dx = we 1 x sinh(2 x) + + C 4 2 Proof. 1 dx 2 1 1 1 × sinh(2 x) − x + C 2 2 2 1 x = sinh(2 x) − + C . 4 2 = (iii) ∫ sech 2 x dx = tanh x + C (iv) ∫ csch 2 x dx = − coth x + C . (v) ∫ tanh 2 x dx = x − tanh x + C (vi) ∫ coth 2 x dx = − coth x + x + C 2 2 Proof. Using 1 − tanh x = sech x , we have Proof. 2 2 ∫ tanh x dx = ∫ (1 − sech x) dx = x − tanh x + C. 5. Cubes of Standard Trig Functions: cos3 x (i) ∫ sin x dx = − cos x + +C 3 sin3 x (ii) ∫ cos x dx = sin x − +C 3 Proof. Proof. 3 3 3 2 ∫ sin x dx = ∫ sin x × sin x dx = ∫ sin x × (1 − cos2 x ) dx = ∫ sin x dx − ∫ sin x × (cos 2 x ) dx = − cos x + cos3 x +C 3 (The last integral is by int. by sub. using u = cos x .) (iii) ∫ tan 3 x dx = tan 2 x + ln cos x + C 2 ∫ sec3 x dx (iv) = Proof. 3 2 ∫ tan x dx = ∫ tan x × tan x dx = ∫ tan x × (sec2 x − 1) dx = ∫ tan x sec 2 x dx − ∫ tan x dx 2 = tan x + ln cos x + C 2 (The second to last integral is by int. by sub. using u = tan x .) 1 1 sec x tan x + ln sec x + tan x + C 2 2 Proof. By int. by parts: u = sec x u′ = sec x tan x v ′ = sec 2 x dx v = tan x ∫ sec3 x dx = u v − ∫ u′v = sec x tan x − ∫ sec x tan 2 x dx = sec x tan x − ∫ sec x (sec 2 x − 1) dx = sec x tan x − ∫ sec3 x dx + ∫ sec x dx So, 2 ∫ sec x dx = sec x tan x + ∫ sec x dx 3 and 1 1 3 ∫ sec x dx = 2 sec x tan x + 2 ∫ sec x dx = 1 1 sec x tan x + ln sec x + tan x + C 2 2