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Power Substitutions

sinm x cosn x m or n odd m or n even u = cos x sin2 = 1 (1 cos 2) 2 cos2 = 1 (1 + cos 2) 2 sin mx cos nx
1 sin A sin B = 2 (cos(A B) cos(A + B))

cos A cos B = 1 (cos(A B) + cos(A + B)) 2 sin A cos B = 1 (sin(A + B) + sin(A B) 2 tanm x secn x m odd u = tan x n even u = sec x
1 sin2 = 2 (1 cos 2) 1 cos2 = 2 (1 + cos 2)

Square Root Substitutions


Expression in Integral a2 x2 a2 + x2 x2 a2 Substitution x = a sin x = a tan x = a sec Identity 1 sin2 = cos2 1 + tan2 = sec2 sec2 1 = tan2

Partial Fractions
P (x) Q(x) Where P (x) and Q(x) are polynomials. If degree of P (x) degree of Q(x), then divide rst.

Case I: Q(x) is a product of unique linear factors


Method: P (x) A1 A2 Ak = + + ... + Q(x) a1 x + b1 a2 x + b2 ak x + bk (1)

Case II: Q(x) factors as a product of (possibly) repeated linear factors.


Method: For each factor (ai x + bi )ri P (x) A1 A2 Ar i = + + ... + Q(x) ai x + bi (ai x + bi )2 (ai x + bi )ri

(2)

Case III: Q(x) factors as a product of unique irreducible quadratic factors.


ie. (b2 4ai ci ) < 0, i : Z i Method: P (x) A1 x + B1 A2 x + B 2 Ak x + B k = + + ... + Q(x) a1 x2 + b1 x + c1 a2 x2 + b2 x + c ak x2 + bk x + ck (3)

Case IV: Q(x) is a product of repeated irreducible quadratic factors.


ie. (b2 4ai ci ) < 0, i : Z i Method: For each factor (ai x + bi )ri P (x) A1 x + B 1 Ari x + Bri Ak x + B k = + + ... + Q(x) ai x2 + bi x + ci (ai x2 + bi x + ci )2 (ai x2 + bi x + ci )ri

(4)

Common Integrals
xn+1 n+1 csc x dx = ln | csc x cot x|

xn dx =

1 dx = ln |x| x

tan x dx = ln | sec x|

cot x dx = ln | sin x| ex dx = ex

ax dx =

ax ln a

sinh x dx = cosh x

cosh x dx = sinh x sin x dx = cos x

cos x dx = sin x

1 x dx dx = tan1 x2 a2 a a

sec2 x dx = tan x

x2

x dx dx = sin1 2 a a

csc2 x dx = cot x

sec x tan x dx = sec x

csc x cot x dx = csc x

sec x dx = ln | sec x + tan x|

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