2020 MathSoc Integration Bee Qualifiers Solutions 1. Standard integral, but the bounds are tricky: Z 507 1 x dx = 5072 − 5032 503 2 1 = (507 + 503) (507 − 503) 2 1 = · 1010 · 4 2 = 2020: 2. Rewrite 3ln x as xln 3 using log laws, so now we have a standard integral: Z Z x1+ln 3 3ln x dx = xln 3 dx = + C: 1 + ln 3 3. This is simply integrating ex=2: Z p Z p ex dx = ex=2 dx = 2 ex + C: p 4. Substitute u = ex − 1: Z p Z u2 Z 1 ex − 1 dx = 2 du = 2 1 − du u2 + 1 1 + u2 = 2 u − tan−1 u + C p p = 2 ex − 1 − tan−1 ex − 1 + C: 5. Substitute x = eu then the integral becomes Z eu cos u du: Apply integration by parts twice: Z Z eu cos u du = eu cos u + eu sin u du Z = eu cos u + eu sin u − eu cos u du: Rearrange to obtain Z 1 eu cos u du = eu (cos u + sin u) ; 2 then Z 1 cos (ln x) dx = x (cos (ln x) + sin (ln x)) + C: 2 6. Substitute u = e2x, then Z e2x 1 Z 1 1 1 p dx = p du = sin−1 u + C = sin−1 e2x + C: 1 − e4x 2 1 − u2 2 2 Z 7π=4 4x cos x 7. This is an odd function from "−a to a" so 2 dx = 0. −7π=4 x − sin jxj + cos jxj c UNSW Mathematics Society 2020 π 8. Substitute u = − x, then 2 Z π=2 sink x Z π=2 cosk u dx = du: k k k k 0 sin x + cos x 0 cos u + sin u Adding the two integrals, we have Z π=2 sink x Z π=2 sink x + cosk x Z π=2 π 2 dx = dx = dx = k k k k 0 sin x + cos x 0 sin x + cos x 0 2 Z π=2 sink x π dx = : k k 0 sin x + cos x 4 9. Multiply numerator and denominator by ex: Z ex dx: 1 + e2x Now substitute u = ex: Z ex Z 1 dx = du = tan−1 u + C = tan−1 (ex) + C: 1 + e2x 1 + u2 10. First, substitute u = tan x: Z Z sec2 (x) sec2 (tan (x)) sec2 (tan (tan (x))) dx = sec2 (u) sec2 (tan (u)) du: Now substitute v = tan (u): Z Z sec2 (u) sec2 (tan (u)) du: = sec2 v dv = tan v + C = tan (tan (tan x)) + C: jxk 11. From 0 to 1, the integrand is identically 0. From 1 to 2, is 0 and bxc is 1. Hence 2 Z 2 jxk Z 2 bxc − 2 dx = dx = 1: 0 2 1 12. Substitute u = 2x, then p p Z 3=4 2x sin−1 (2x) 1 Z 3=2 u sin−1 u p dx = p du: 2 2 0 1 − 4x 2 0 1 − u Now substitute u = sin v: p 1 Z 3=2 u sin−1 u 1 Z π=3 p du = v sin v dv: 2 2 0 1 − u 2 0 Here, we use by parts to obtain the answer, p 1 Z π=3 3 π v sin v dv = − : 2 0 4 12 13. Substitute x = u2, then Z 1 p Z 1 sin−1 x dx = 2u sin−1 u du: 0 0 Now substitute u = sin v, then Z 1 Z π=2 2u sin−1 u du = v sin 2v dv: 0 0 Here, we use by parts to obtain the final answer: Z π=2 π v sin 2v dv = : 0 4 c UNSW Mathematics Society 2020 14. First we multiply the numerator and denominator by cos4 x: Z π=2 1 Z π=2 cos4 x dx = dx: 4 4 4 0 1 + tan x 0 cos x + sin x Now we can use the answer from Q8: Z π=2 cos4 x π dx = : 4 4 0 cos x + sin x 4 15. From 0 to π=2 the integrand is x. From π=2 to 3π=2 the integrand is π − x. Hence Z 3π=2 Z π=2 Z 3π=2 sin−1 (sin x) dx = x dx + (π − x) dx 0 0 π=2 π2 π2 = + π2 − π2 = : 8 8 16. Separate the integrand via partial fractions: Z 1 dx Z 1 1 x 2 = − 2 dx: 1 x(x + 1) 1 x x + 1 These can both be integrated into logarithms: Z 1 1 x h p i1 x 1 1 − dx = ln x − ln 1 + x2 = ln p = ln 2: 2 2 1 x x + 1 1 1 + x 1 2 17. Multiply the numerator and denominator by 1 − sin x: Z π=2 Z π=2 Z π=2 1 1 − sin x 2 sin x dx = 2 dx = sec x − 2 dx 0 1 + sin x 0 1 − sin x 0 cos x π=2 sin x − 1 = [tan x − sec x]0 = lim + 1 x!π=2 cos x = 1: 18. Substitute x = cos u, then Z −1 Z ecos x dx = − eu sin u du: We can apply integration by parts here, or simply quote the result: Z 1 eu sin u du = eu (sin u − cos u) : 2 Hence Z −1 1 −1 ecos x dx = − ecos x sin cos−1 x − cos cos−1 x + C: 2 p We can simplify further by observing that sin cos−1 x = 1 − x2 and cos cos−1 x = x so Z −1 1 −1 p ecos x dx = ecos x x − 1 − x2 + C: 2 19. Begin by transforming the denominator into a cosine function via auxiliary angle method: 4 3 cos x + 4 sin x = 5 cos x − sin−1 : 5 c UNSW Mathematics Society 2020 Substituting this result into the integral: Z π=2 Z π=2 25 2 −1 4 2 dx = sec x − sin dx 0 (3 cos x + 4 sin x) 0 5 4π=2 = tan x − sin−1 5 0 π 4 4 = tan − sin−1 + tan sin−1 2 5 5 4 4 = cot sin−1 + tan sin−1 5 5 3 4 25 = + = : 4 3 12 20. Multiply the numerator and denominator by x−7: Z 1 Z x−7 1 dx = dx = − ln 1 + x−6 + C: x7 + x 1 + x−6 6 c UNSW Mathematics Society 2020 2020 MathSoc Integration Bee Team Standoff Solutions • Team A Question 1: Substitute u = ln x, then Z x − 1 Z eu − 1 dx = du: x + x2 ln x 1 + ueu Now we multiply numerator and denominator by e−u: Z u Z −u e − 1 −e + 1 −u du = du = ln e + u + C: 1 + ueu e−u + u • Team A Question 2: We have symmetry around x = 1 so Z 2 π jx − 1j Z 1 π jx − 1j sin2 dx = 2 sin2 dx: 0 2 0 2 For this interval, jx − 1j = 1 − x so Z 1 π jx − 1j Z 1 π(1 − x) 2 sin2 dx = 2 sin2 dx 0 2 0 2 Z 1 πx = 2 cos2 dx 0 2 Z 1 = (1 − cos (πx)) dx 0 = 1: • Team A Question 3: Substitute x = u2: Z 1=4 p Z 1=2 e x dx = 2ueu du: 0 0 Then by integration by parts, Z 1=2 Z 1=2 u u 1=2 u 2ue du = 2 [ue ]0 − 2 e du 0 p 0 = 2 − e: • Team B Question 1: First we complete the square in the square root, s p 1 12 x − x2 = − x − : 4 2 Hence our integral can be written as Z 1 Z 1 p dx = dx 2 s x − x 1 12 − x − 4 2 Z 1 = 2 q dx 1 − (2x − 1)2 = sin−1 (2x − 1) + C: c UNSW Mathematics Society 2020 π • Team B Question 2: Using the substitution u = − x, 4 Z π=4 Z π=4 π ln (1 + tan x) dx = ln 1 + tan − x dx: 0 0 4 π 1 − tan x However tan − x = so 4 1 + tan x Z π=4 Z π=4 1 − tan x ln (1 + tan x) dx = ln 1 + dx 0 0 1 + tan x Z π=4 2 = ln dx 0 1 + tan x Z π=4 Z π=4 = ln 2 dx − ln (1 + tan x) dx 0 0 Z π=4 π ln 2 ln (1 + tan x) dx = : 0 8 π • Team B Question 3: We use the substitution u = − x: 2 Z π=2 cos2 x Z π=2 sin2 x dx = : 0 sin x + cos x 0 cos x + sin x Hence by adding the two integrals, we have Z π=2 cos2 x Z π=2 sin2 x + cos2 x Z π=2 1 2 dx = dx = dx: 0 sin x + cos x 0 sin x + cos x 0 sin x + cos x Now use the auxiliary angle method on the denominator: p π sin x + cos x = 2 cos x − : 4 So the RHS integral becomes Z π=2 1 Z π=2 1 π dx = p sec x − dx 0 sin x + cos x 0 2 4 1 π π π=2 = p ln sec x − + tan x − 2 4 4 0 p ! 1 2 + 1 = p ln p : 2 2 − 1 Hence our original integral is p ! Z π=2 cos2 x 1 2 + 1 dx = p ln p : 0 sin x + cos x 2 2 2 − 1 • Team C Question 1: First we divide the numerator and denominator by x2: Z x2 − 1 Z 1 − x−2 dx = dx: x4 + 1 x2 + x−2 1 Now by using the substitution u = x + , x Z 1 − x−2 Z 1 dx = du: x2 + x−2 u2 − 2 c UNSW Mathematics Society 2020 Separating the integrand using partial fractions, p Z 1 1 Z 1 1 1 u − 2 du = p p − p du = p ln p + C: u2 − 2 2 2 u − 2 u + 2 2 2 u + 2 Hence our integral is p Z x2 − 1 1 x2 − 2x + 1 dx = p ln p + C: x4 + 1 2 2 x2 + 2x + 1 • Team C Question 2: The terms of odd power integrate to 0, so we only need to consider the even powered terms.