Formula Sheet for Math 175 Integrals Z Z tan x dx = ln | sec x| + C cot x dx = − ln | csc x| + C Z Z sec x dx = ln | sec x + tan x| + C Z csc x dx = − ln | csc x + cot x| + C Z sec2 x dx = tan x + C Z Z csc x cot x dx = − csc x + C sec x tan x dx = sec x + C Z csc2 x dx = − cot x + C √ Z x dx = sin−1 +C a a2 − x2 Trigonometric Substitution p a2 − x2 −→ x = a sin θ, p dx 1 −1 x = tan +C a2 + x2 a a a2 + x2 −→ x = a tan θ, p x2 − a2 −→ x = a sec θ Trigonometric Identities 1 + tan2 θ = sec2 θ, cos2 θ = 1 + cos 2θ , 2 sin2 θ = 1 + cot2 θ = csc2 θ 1 − cos 2θ , 2 sin 2θ = 2 sin θ cos θ Series f 00 (a) f 000 (a) (x − a)2 + (x − a)3 + · · · 2! 3! k(k − 1)x2 k(k − 1)(k − 2)x3 (1 + x)k = 1 + kx + + + · · · , −1 < x < 1 2! 3! a a + ar + ar2 + ar3 + · · · = , −1 < r < 1 1−r x3 x5 x7 x2 x4 x6 x2 x3 x4 sin x = x − + − + · · · , cos x = 1 − + − + · · · , ex = 1 + x + + + +··· 3! 5! 7! 2! 4! 6! 2! 3! 4! f (x) = f (a) + f 0 (a)(x − a) + Differential Equations dy + P (x)y = Q(x) dx R Integrating factor: e P (x) dx ay 00 (x) + by 0 (x) + cy(x) = 0 y = c1 em1 x + c2 em2 x y = (c1 + c2 x)emx y = eαx (c1 cos(βx) + c2 sin(βx)) Fourier Series If f (x) has period T = 2L, then f (x) = a0 + ∞ X ak cos k=1 1 a0 = 2L Z L f (x) dx, −L 1 ak = L Z L f (x) cos −L kπx L kπx L dx, + ∞ X bk sin k=1 1 bk = L Z kπx L L f (x) sin −L kπx L dx, k = 1, 2, 3 . . . f (t) L {f (t)} = F (s) f (t) 1 1 s eat 1 s−a eat tn n! 1 (s − a)n+1 n! tn sn+1 L {f (t)} = F (s) sin ωt ω s2 + ω 2 cos ωt s s2 + ω 2 eat sin ωt ω (s − a)2 + ω 2 eat cos ωt s−a (s − a)2 + ω 2 t sin ωt 2ωs (s2 + ω 2 )2 t cos ωt s2 − ω 2 (s2 + ω 2 )2 sin ωt − ωt cos ωt 2ω 3 1 (s2 + ω 2 )2 U(t − a) e−as s δ(t) 1 δ(t − a) e−as L {αf (t) + βg(t)} = αL {f (t)} + βL {g(t)} L {eat f (t)} = F (s − a) L {tn f (t)} = (−1)n L −1 {e−as F (s)} = f (t − a) U(t − a) L {f 0 (t)} = sF (s) − f (0) dn F (s) dsn L {f (t) U(t − a)} = e−as L {f (t + a)} L {f 00 (t)} = s2 F (s) − sf (0) − f 0 (0) L {f (n) (t)} = sn F (s) − sn−1 f (0) − · · · − f (n−1) (0) Z f (t) ∗ g(t) = t f (θ)g(t − θ) dθ =⇒ L {f (t) ∗ g(t)} = F (s)G(s) 0 L Z t f (θ) dθ = 0 f (t) has period T =⇒ L {f (t)} = F (s) s 1 1 − e−sT Z 0 T e−st f (t) dt