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3450:335 Ordinary Differential Equations Review of Integration and Trigonometry Here is a review of the integration and trigonometric skills you need in this course. Integration by substitution Z Z p e2x dx = 2x3 1 + x2 dx = 1 2x e + c u = 2x, du = 2dx Z2 Z (u − 1)u1/2 du = u3/2 − u1/2 du u = 1 + x2 , du = 2xdx, x2 = u − 1 Integration by parts There Z are 3 classes of integrals that Z can be evaluated using IBP: A) (polynomial)(sin or cos) dx, polynomial one degree at a time. Z B) (polynomial)(logarithm) dx, (polynomial)(exp) dx. Let (u = polynomial) to kill off the Z arctrig dx. Let u = logarithm or u = arctrig. Z C) (exp)(sin or cos) dx. IBP twice, using u = trig both times, or u = exp both times. Called FOLDING. Basic Trigonometric Skills A) Definitions of the 6 trig functions in terms of right triangles and in terms of sines and cosines. B) Pythagorean identities, 3 flavours. sin2 x+cos2 x = 1, 1+cot2 x = csc2 x, tan2 x+1 = sec2 x. C) Half angle identities. sin2 x = 21 (1 − cos(2x)), cos2 x = 12 (1 + cos(2x)). D) Double angle identities. 2 sin x cos x = sin(2x), cos2 x − sin2 x = cos(2x), 2 cos2 x − 1 = cos(2x), 1 − 2 sin2 x = cos(2x). E) Sum and difference identities. sin(x ± y) = sin x cos y ± sin y cos x, cos(x ± y) = cos x cos y ∓ sin x sin y. F) Product identities. sin A cos B = 12 (sin(A − B) + sin(A + B)), sin A sin B = 12 (cos(A − B) − cos(A + B)), cos A cos B =R 12 (cos(A − B) + cos(A + RB)). R G) Elementary antiderivatives. R sin x dx = cos x + c, cos x dx = − sin x + c, sec2 x dx = R 2 tan x + c, csc x dx = − cot x + c, sec x tan x Z dx = sec x +Z c. Z 1 sin x H) Antiderivatives using logarithmic form. tan x dx = dx = − du = − ln | cos x|+ cos x u Z Z Z 1 sec x + tan x c, sec x dx = sec x dx = du = ln | sec x + tan x| + c. sec x + tan x u Trigonometric Integrals Z A) sinm x cosn x dx. If n or m is odd, pull off the odd power of sin or cos to use in du, let u equal the other trig function, and apply Pythagorean identities if needed. If both n and m are odd, you can let u equal sin x or cos x. If n and m are both even, apply half angle identities and Pythagorean identities. 1 Examples: Z Z 3 cos x dx = Z sin3 x cos3 x dx = Z Z 2 (1 − sin x)(cos xdx) = Z Z (u3 )(1 − u2 )(du) 1 (1 − cos(2x)) dx 2 sin x dx = B) (1 − u2 )du (sin3 x)(1 − sin2 x)(cos xdx) = Z 2 Z tanm x secn x dx. If n is even, pull off a sec2 x to use in du, let u = tan x and apply Pythagorean identity if needed. If m is odd, pull off a sec x tan x to use in du, let u = sec x, apply Pythagorean identity if needed. If both n is even and m is odd, you can do either of the above. If n is odd and m is even, do your best. Use Pythagorean identities to reduce to nothing but powers of sec x. Integrals of sec x and sec3 x are standard, and can be looked up in your old Calculus text (you didn’t sell it back to the Bookstore, did you?). Higher odd powers of sec x require careful use of IBP. Even powers can be handled by the second sentence in this paragraph. Examples: Z 2 4 Z tan x sec x dx = Z Z 2 2 2 tan x(tan x − 1)(sec x dx) = tan x sec20 x dx = Z sec19 x(sec x tan x dx) = tan2 x sec x dx = Z (sec2 x − 1) sec x dx = Z Z Z u2 (u2 − 1) du u19 du sec3 x − sec x dx Trigonometric Substitution √ √ √ Look for expressions of the form a2 − x2 , x2 − a2 or x2 + a2 , or one of these to a higher power (for example, (x2 − 4)2 is the square root raised to the fourth power). If there is a minus under the square root, the positive term is the square of the hypotenuse and the negative term is the square of one of the legs; the square root is the other leg. If there is a plus under the square root, then the square root is the hypotenuse, and the terms inside are the squares of the legs. Use the triangle to make a substitution, which converts the integral to a trig integral (see above). Examples: Z √ 1 √ 1) dx. The hypotenuse is 4 and the sides are x and 16 − x2 . Let x = 4 sin θ, x2 16 − x2 Z Z 1 1 p 4 cos θ dθ = 1 sin2 θ dθ = so dx = 4 cos θ. The integral becomes 2 2 16 16 sin θ 16 − 16 sin θ √ 1 16 − x2 1 − cot θ + c = − + c. x 16 Z 16 √ x3 √ dx. The hypotenuse is 9 + x2 and the legs are x and 3. Let x = 3 tan θ, so 2) 9 + x2 Z Z 27 tan3 θ sin3 θ 2 2 dx = 3 sec θdθ. The integral becomes 3 sec θ dθ = 27 dθ. Write the numerator 3 sec θ cos4 θ Z as sin2 θ(sin θ dθ) = (1 − u2 )du with u = cos θ. The integral becomes 27 (1 − u2 )u−4 (−du) = p √ 1 1 1 27 − + 3 = −9 9 + x2 + (9 + x2 )3/2 . Note that u = cos θ = 3/ 9 + x2 . Oh, and u 3u 3 don’t forget the constant of integration. 2