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1. Trigonometric Identities 2 The Pythagorean Theorem, sin x + cos2 x = 1, has other forms, like (1) sin2 x cos2 x 1 + = ⇒ tan2 x + 1 = sec2 x 2 2 cos x cos x cos2 x and (2) sin2 x cos2 x 1 + = ⇒ 1 + cot2 x = csc2 x. 2 2 2 sin x sin x sin x Angle Addition: (3) sin(α + β) = sin α cos β + cos α sin β (4) cos(α + β) = cos α cos β − sin α sin β The angle addition formulae give us the double angle formulae: (5) sin(2θ) = 2 sin θ cos θ (6) cos(2θ) = cos2 θ − sin2 θ Because cos(2θ) = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1, we have the half-angle formula cos2 θ = (7) 1 + cos(2θ) . 2 Similarly, because cos(2θ) = 1 − sin2 θ − sin2 θ, we find the other half-angle formula sin2 θ = (8) 1 − cos(2θ) . 2 2. Derivatives of Trigonometric Functions We know the derivatives d (9) sin x = cos x dx which allow us to calculate and d cos x = − sin x, dx (10) d cos x sin x cos2 x + sin2 x tan x = + sin x 2 = = sec2 x dx cos x cos x cos2 x (11) d − cos x csc x = = − csc x cot x dx sin2 x (12) d sin x sec x = = sec x tan x dx cos2 x (13) − sin2 x − cos2 x d − sin x − cos x = = − csc2 x cot x = + cos x dx sin x sin2 x sin2 x 1 2 To compute R 3. Useful Integrals cos2 xdx, (14) Similarly, to compute R (15) To compute R To compute R sin2 xdx, we use the half-angle formula for sines: Z Z 1 − cos(2x) x sin(2x) 2 sin xdx = dx = − +C 2 2 4 dx , 1−x2 Z (16) we use the half-angle formula for cosines: Z Z 1 + cos(2x) x sin(2x) cos2 xdx = dx = + +C 2 2 4 we use the method of partial fractions: ¶ Z µ dx 1 1 1 1 = + dx = [ln(1 + x) − ln(1 − x)] + C 2 1−x 2 1+x 1−x 2 4. Integrals using Trigonometric Substitutions dx , 1+x2 we substitute x = tan θ ⇒ dx = sec2 θdθ (17) to find Z (18) To compute R √ dx , 1−x2 dx = 1 + x2 (20) sec2 θ dθ = θ + C = arctan x + C sec2 θ we substitute (19) to find Z x = sin θ ⇒ dx = cos θdθ Z dx √ = 1 − x2 Z cos θ p 1 − sin2 θ dθ = θ + C = arcsin x + C 5. Integration by Parts The Product Rule states that d(uv) = udv + vdu. Integrating, we find Z Z Z Z Z Z Z (21) d(uv) = udv + vdu ⇒ uv = udv + vdu ⇒ udv = uv − vdu. For example, to integrate xex , we choose u = x and v = ex , so that du = dx and dv = ex , and calculate Z Z 0 (22) xex dx = xex + C − ex dx = xex − ex + C