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SUMMARY OF INTEGRATION TECHNIQUES Integration by Standard forms: xn+1 1 (Ax + B)n+1 [f(x)]n+1 n n n dx = + c, (Ax + B) dx = + c, f '(x) [f(x)] dx = x n+1 n + 1 + c, A n+1 for n –1 1 x dx = ln | x | + c for n –1 for n –1 1 f '(x) 1 f(x) dx = ln | f(x) | + c Ax + B dx = A ln | Ax + B | + c 1 Ax+B f '(x) e f(x) dx = e f(x) + c ex dx = ex + c dx = A eAx+B + c e 1 aAx+B af(x) ax x Ax+B f(x) a dx = + c a dx = + c f '(x) a dx = ln a ln a + c A ln a Integration of Trigonometric Functions: f (x) sin f(x) dx = –cos f(x) + c sin x dx = –cos x + c cos x dx = sin x + c 2 sec x dx = tan x + c f (x) cos f(x) dx = sin f(x) + c 2 f (x) sec f(x) dx = tan f(x) + c cosec2 x dx = –cot x + c sec x tan x dx = sec x + c f (x) cosec2 f(x) dx = –cot f(x) + c f (x) sec f(x) tan f(x) dx = sec f(x) + c cosec x cot x dx = –cosec x + c 2 2 tan x dx = sec x – 1 dx = tan x – x + c f (x) cosec f(x) cot f(x) dx = –cosec f(x) + c 2 cot2 x dx = cosec x – 1 dx = – cot x – x + c Using Double Angle Formulas: cos2 x dx 1 + cos 2x x sin 2x dx =2+ 4 +c 2 1 – cos 2x x sin 2x = dx =2– 4 +c 2 1 = sin (a + b)x + sin (a – b)x dx 2 1 = sin (a + b)x – sin (a – b)x dx 2 1 = cos (a + b)x + cos (a – b)x dx 2 1 = –2 cos (a + b)x – cos (a – b)x dx 1 –1 x a tan a + c x sin–1 a + c , |x|<a 1 x–a x>a 2a ln x + a + c , 1 a+x ln |x|<a 2a a – x + c, ln (sec x) + c , |x|<2 ln (sin x) + c , 0<x< –ln (cosec x + cot x) + c , 0<x< ln (sec x + tan x) + c , |x|<2 uv– u v dx = 2 sin x dx Using Factor Formulas: sin ax cos bx dx cos ax sin bx dx cos ax cos bx dx sin ax sin bx dx Given in formula list: 1 x2 + a2 dx = 1 a2 – x2 dx = 1 x2 – a2 dx = 1 a2 – x2 dx = tan x dx = = cot x dx cosec x dx = sec x dx = = u v dx Order for choosing u: Logarithm, Inverse Trigonometric, Algebraic, Trigonometric, Exponential Integration by Parts: 12 E.g. e ln x dx 1 E.g. 1 Let u = ln x u = x Let v = 1 v = x e e 1 = [ x ln x ]1 – 1 x x dx e = [ e ln e ] – 1 1 dx e = e – [ x ]1 =e–[e–1] =1 E.g. ex sin x dx x = ex sin x – e cos x dx x = ex sin x – [ ex cos x – e (–sin x) dx ] x x x 2 e sin x dx = e sin x – e cos x 1 x x e sin x dx = 2 e (sin x – cos x) + c 2 d 2 3 x Find dx ex and deduce x e dx. d x2 x2 dx e = 2x e 2 3 x x e dx 2 1 2 x2 =2 Let u = x2, v = 2xex x 2x e dx 2 1 x2 = 2 [ x2 ex – 2x e dx ] 2 2 1 = 2 [ x2 ex – ex ] + c Let u = sin x, v'= ex u' = cos x, v = ex Let u = cos x, v'= ex u' = –sin x, v = ex Integration by Splitting the Numerator 4x + 5 E.g. x2 + 2x + 3 dx f (x) 2x + 2 1 1 =2 rewrite the integral as x2 + 2x + 3 dx + (x + 1)2 + 2 dx f(x) dx + quadratic dx 1 x+1 = 2 ln | x2 + 2x + 3 | + tan–1 +c 2 2 Integration by Partial Fractions 3x3 + x 3 2 1 E.g. break up into partial fractions (x + 1)2(x2 + 1) dx = x + 1 – (x + 1)2 – x2 + 1 dx 2 = 3 ln | x + 1 | + x + 1 – tan–1 x + c Integration by Substitution 1 1 E.g. Use the substitution x = tan to solve 2 0 x + 1 dx dx x = tan = sec2 use this to replace dx d When x = 0, tan = 0 = 0 to replace the two limits When x = 1, tan = 1 = 4 1 1 0 x2 + 1 dx 1 1 /4 = sec2 d substitute the integrand 2 , dx and the two limits 2 0 x +1 tan + 1 /4 = 0 sec d /4 = [ ln | sec + tan | ] 0 = ln ( 2 + 1 ) 13