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Math 1B Spring07

Math 1B Spring07 - Techniques of Integration(Ch.7...

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Techniques of Integration (Ch.7) Integration by Parts (a) ∫udv = uv - ∫vdu (b) ∫ a b f(x)g’(x)dx = f(x)g(x)] a b - ∫ a b g(x)f’(x)dx Trigonometric Integration Evaluating ∫sin m x cos n x dx (a) Power of cosine is odd (n = 2k + 1), save one cosine factor and use cos 2 x = 1 – sin 2 x to express remaining factors in terms of sine: ∫sin m x cos 2k+1 x dx = ∫sin m x (cos 2 x) k cos x dx = ∫sin m x (1 - sin 2 x) k cos 2 x dx substitute u = sin x (b) Power of sine is odd (m = 2k + 1), save one sine factor and use sin 2 x = 1 – cos 2 x to express remaining factors in terms of cosine: ∫sin 2k+1 x cos n x dx = ∫(sin 2 x) k cos n x sin x dx = ∫(1 - cos 2 x) k cos n x sin x dx substitute u = cos x (c) Power of sine and cosine are odd either 1) or 2) may be used (d) Power of sine and cosine are both even, use the half angle-identities sin 2 x = ½(1- cos2x) cos 2 x = ½(1+ cos2x) Evaluating ∫tan m x sec n x dx (a) Power of secant is even (n = 2k, k ≥ 2), save a factor of sec 2 x and use sec 2 x = 1 + tan 2 x to express remaining factors in terms of tan x: ∫tan m x sec 2k x dx = ∫tan m x (sec 2 x) k-1 sec 2 x dx = ∫tan m x (1 + tan 2 x) k-1 sec 2 x dx substitute u = tan x (b) Power of tangent is odd (m = 2k + 1), save a factor of sec x tan x and use tan 2 x = sec 2 x – 1 to express remaining factors in terms of sec x: ∫tan 2k+1 x sec n x dx = ∫(tan 2 x) k sec n-1 x sec x tan x dx = ∫(sec 2 x – 1) k sec n-1 x sec x tan x dx substitute u = sec x (c) May have to use the formulas: ∫tan x dx = ln │sec x│ + C ∫sec x dx = ln │sec x + tan x│ + C Evaluating the Integrals a) ∫sin mx cos nx dx, b) ∫sin mx sin nx dx, or c) ∫cos mx cos nx dx (a) sin A cos B = ½ [sin (A – B) + sin (A + B)] (b) sin A sin B = ½ [cos (A – B) - cos (A + B) ] (c) cos A cos B = ½ [cos (A – B) + cos (A + B)] Trigonometric Substitution Expression Substitution Identity √(a 2 – x 2 ) x = a sin θ 1 – sin 2 θ = cos 2 θ √(a 2 + x 2 ) x = a tan θ 1 + tan 2 θ = sec 2 θ √(x 2 – a 2 ) x = a sec θ sec 2 θ – 1 = tan 2 θ Improper Integration Type I (a) ∫ a f(x)dx = lim t→∞ a t f(x)dx (b) ∫ -∞ b f(x)dx = lim t→-∞ t b f(x)dx - if the limit exists then the improper integral is convergent - if the limit doesn’t exist then the improper integral is divergent (c) ∫ -∞ f(x)dx = ∫ -∞ a f(x)dx + ∫ a f(x)dx - but only if both integrals are convergent Type II (a) If f is continuous on [a,b) and discontinuous at b, then a b f(x)dx = lim t→b - a t f(x)dx (b) If f is continuous on (a,b] and discontinuous at a, then a b f(x)dx = lim t→a - t b f(x)dx
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