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Summer 2016 MTH142J College Calculus 2 Trigonometric integration Case Power of sin x Power of cos x Solution (1) Odd > 0 Any Substitution u = cos x, du = − sin x d x. Use sin2 x = 1 − cos2 x to transform sine to cosine. (2) Any Odd > 0 (3) Even ≥ 0 Any (4) Any Even ≥ 0 Substitution u = sin x, du = cos x d x. Use cos2 x = 1 − sin2 x to transform cosine to sine. R Use sin2 x = 1 − cos2 x to transform all sine to cosine. Then use reduction formula cosn x d x. R Use cos2 x = 1 − sin2 x to transform all cosine to sine. Then use reduction formula sinn x d x. Z Z tan x d x = ln |sec x| + C, sec x d x = ln |sec x + tan x| + C Case Power of tan x Power of sec x Solution (5) Odd > 0 Any Substitution u = sec x, du = sec x tan x d x. Use tan2 x = sec 2 x − 1 to transform tangent to secant. (6) Any Even > 0 (7) Even ≥ 0 Any Substitution u = tan x, du = sec 2 x d x. Use sec 2 x = 1 + tan2 x to transform secant to tangent. R Use tan2 x = sec 2 x − 1 to transform all tangent to secant. Then use reduction formula sec n x d x. Reduction formulas for integrals of trigonometric power: Z Z 1 n−1 n−1 sin x d x = − sin x · cos x + sinn−2 x d x n n Z Z 1 n−1 cosn x d x = cosn−1 x · sin x + cosn−2 x d x n n Z Z 1 tann−1 x − tann−2 x d x tann x d x = n−1 Z Z 1 n−2 n−2 csc x d x = − csc x · cot x + csc n−2 x d x n−1 n−1 Z Z 1 n−2 sec n x d x = sec n−2 x · tan x + sec n−2 x d x n−1 n−1 Z Z 1 cotn x d x = − cotn−1 x − cotn−2 x d x n−1 n n Trigonometric values of typical angles: π 6 = 45◦ π 3 0 sin θ 0 1 2 p 2 2 cos θ 1 p 3 2 p 2 2 1 2 tan θ 0 p 3 3 1 p 3 1 p 2 3 3 p 2 sec θ = 30◦ π 4 θ = 60◦ p 3 2 2 π 2 = 90◦ 2π 3 = 120◦ 3π 4 = 135◦ 5π 6 = 150◦ π = 180◦ p 3 2 p 2 2 1 2 0 0 − 12 p − 22 p − 23 −1 DNE p − 3 −1 −2 p − 2 1 DNE − p 3 3 p − 233 0 −1 Trigonometric substitution Form Substitution Differential After transformation a2 − x 2 x = a sin θ d x = a cos θ dθ a2 − x 2 = a2 cos2 θ x 2 + a2 x = a tan θ d x = asec 2 θ dθ x 2 + a2 = a2 sec 2 θ x 2 − a2 x = asec θ d x = asec θ tan θ dθ x 2 − a2 = a2 tan2 θ Definition of Trigonometric functions: OP HP AJ cos θ = HP OP sin θ tan θ = = AJ cos θ sin θ = reciprocal ←→ reciprocal ←→ reciprocal ←→ HP OP HP sec θ = AJ cos θ AJ = cot θ = OP sin θ csc θ = Pythagorean theorem: OP2 + AJ2 = HP2 . Double angle identities sin2 θ = 1 − cos 2θ , 2 cos2 θ = 1 + cos 2θ 2 p a2 − x 2 = a cos θ p x 2 + a2 = asec θ p x 2 − a2 = a tan θ