2020 Mathsoc Integration Bee Qualifiers Solutions

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 √ Z √ ex dx = ex/2 dx = 2 ex + C.

√ 4. Substitute u = ex − 1:

Z √ Z u2 Z 1 ex − 1 dx = 2 du = 2 1 − du u2 + 1 1 + u2 = 2 u − tan−1 u + C √ √ = 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 √ dx = √ 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 |x| + cos |x|

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 √ √ Z 3/4 2x sin−1 (2x) 1 Z 3/2 u sin−1 u √ dx = √ du. 2 2 0 1 − 4x 2 0 1 − u Now substitute u = sin v: √ 1 Z 3/2 u sin−1 u 1 Z π/3 √ du = v sin v dv. 2 2 0 1 − u 2 0 Here, we use by parts to obtain the answer, √ 1 Z π/3 3 π v sin v dv = − . 2 0 4 12 13. Substitute x = u2, then Z 1 √ 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 ∞ dx Z ∞ 1 x 2 = − 2 dx. 1 x(x + 1) 1 x x + 1 These can both be integrated into logarithms:

Z ∞ 1 x h p i∞ x ∞ 1 − dx = ln x − ln 1 + x2 = ln √ = 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 √ 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 |x − 1| Z 1 π |x − 1| sin2 dx = 2 sin2 dx. 0 2 0 2 For this interval, |x − 1| = 1 − x so

Z 1 π |x − 1| 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 √ 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 √ 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 √ 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: √ π sin x + cos x = 2 cos x − . 4 So the RHS integral becomes

Z π/2 1 Z π/2 1 π dx = √ sec x − dx 0 sin x + cos x 0 2 4 1 π π π/2 = √ ln sec x − + tan x − 2 4 4 0 √ ! 1 2 + 1 = √ ln √ . 2 2 − 1

Hence our original integral is √ ! Z π/2 cos2 x 1 2 + 1 dx = √ ln √ . 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, √ Z 1 1 Z 1 1 1 u − 2 du = √ √ − √ du = √ ln √ + C. u2 − 2 2 2 u − 2 u + 2 2 2 u + 2 Hence our integral is √ Z x2 − 1 1 x2 − 2x + 1 dx = √ ln √ + 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. So 9 Z 1 X Z 1 kxk dx = 2 2x2 + 4x4 + 6x6 + 8x8 dx −1k=0 0 1 2 3 4 = 4 + + + . 3 5 7 9 • Team C Question 3: Using the substitution u = −x, Z 1 Z 1 −1 x −1 1 tan (2 ) dx = tan u du. −1 −1 2 Note that 2x > 0, so adding this integral to the original give us Z 1 Z 1 −1 x 1 −1 x −1 1 tan (2 ) dx = tan (2 ) + tan x dx −1 2 −1 2 1 Z 1 π = dx 2 −1 2 π = . 2 • Team D Question 1: Substitute u = x3/2, then Z r x 2 Z 1 dx = √ du. 1 − x3 3 1 − u2 This is a standard inverse sine integral, so Z r x 2 2 dx = sin−1 u + C = sin−1 x3/2 + C. 1 − x3 3 3

• Team D Question 2: Pull a factor of x4 out of the brackets: Z 1 Z 1 dx = dx. x2 (x4 + 1)3/4 x5 (1 + x−4)3/4 Now by using the substitution u = 1 + x−4, Z 1 1 Z dx = − u−3/4 du x5 (1 + x−4)3/4 4 = −u1/4 + C 1 = − 1 + x−4 4 + C

• Team D Question 3: Using the substitution u = 1 + ln x, 2 Z e ln (1 + ln x) Z 3 dx = ln u du. 1 x 1 Now apply integration by parts: Z 3 Z 3 3 ln u du = [u ln u]1 − du 1 1 = 3 ln (3) − 2.

c UNSW Mathematics Society 2020 2020 MathSoc Integration Bee Semi-Finals Solutions d • Round 1 Question 1: Observe that (x sin x + cos x) = x cos x. We want this in the numerator: dx Z x2 Z x x cos x dx = dx. (x sin x + cos x)2 cos x (x sin x + cos x)2

Now we can use integration by parts to evaluate the integral, Z x x cos x −x sec x Z dx = + sec2 x dx cos x (x sin x + cos x)2 x sin x + cos x −x sec x + sec x sin x (x sin x + cos x) = + C x sin x + cos x sin x − x cos x = + C. x sin x + cos x

• Round 1 Question 2: In order to evaluate this integral, we seek a reduction formula. This can be done via integration by parts. We set In to be the integral, then

m+1 1 Z 1 m+1 n−1 x n x n (ln x) In = (ln x) − dx m + 1 0 0 m + 1 x n Z 1 = 0 − xm (ln x)n−1 dx m + 1 0 n = − I . m + 1 n−1 I Hence by taking the product of the ratios k from k = 1 to k = n: Ik−1

n n Y Ik Y (−1)k = Ik−1 m + 1 k=1 k=1 n In (−1) n! = n . I0 (m + 1)

Now we must evaluate I0, which is a standard integral Z 1 m 1 I0 = x dx = . 0 m + 1 Hence (−1)nn! I = . n (m + 1)n+1

• Round 1 Question 3: Using the substitution u = 808 − x,

Z 405 pln (2020 − x) Z 405 pln (1212 + u) p p dx = p p du. 403 ln (2020 − x) + ln (1212 + x) 403 ln (1212 + u) + ln (2020 − u)

Adding this integral to the original, we have

Z 405 pln (2020 − x) 1 Z 405 pln (2020 − x) + pln (1616 + x) p p dx = p p dx 403 ln (2020 − x) + ln (1212 + x) 2 403 ln (2020 − x) + ln (1212 + x) 1 Z 405 = dx = 1. 2 403

c UNSW Mathematics Society 2020 • Round 2 Question 1: We require a reduction formula, which can be found via integration by parts. Denote the original integral as In, then Z ∞ 1 2n−1 −x2 In = − x −2xe dx 2 0 ∞ 1 h 2 i∞ 1 Z 2 = − x2n−1e−x + (2n − 1)x2n−2e−x dx 2 0 2 0 2n − 1 = I . 2 n−1 I Taking the product of the ratios k from k = 1 to k = n, Ik−1

n n Y Ik Y 2k − 1 = Ik−1 2 k=1 k=1 In (2n − 1)!! = n . I0 2

Evaluating I0: Z ∞ √ −x2 π I0 = e dx = . 0 2 Hence √ (2n − 1)!! π I = . n 2n+1 1 1 1 1 • Round 2 Question 2: For < x < , the integrand is ln 3. Also for < x < , the integrand is π 3 3 e ln 2. So, we are integrating constants along two separate bounds:

Z 1/e 1 Z 1/3 Z 1/e ln dx = ln 3 dx + ln 2 dx. 1/π x 1/π 1/3 1 1 1 1 = − ln 3 + − ln 2. 3 π e 3

• Round 2 Question 3: We need to apply integration by parts twice:

Z 1 Z 1 1 sin (x) sinh (x − 1) dx = [sin (x) cosh (x − 1)]0 − cos (x) cosh (x − 1) dx 0 0 Z 1 1 = sin (1) − [cos (x) sinh (x − 1)]0 − sin (x) sinh (x − 1) dx 0 Z 1 1 sin (x) sinh (x − 1) dx = (sin (1) − sinh (1)) . 0 2

c UNSW Mathematics Society 2020 2020 MathSoc Integration Bee Runner Up Solutions

• Harder Integral: First, notice that the denominator can be simpliﬁed as −1 p 2 p 2 sin tan 1 − x /x = 1 − x . Then

√ r1 + x r1 + x 1 − x2 + sin−1 sin−1 Z 1 2 Z 1 2 √ dx = 1 + √ dx 2 2 0 1 − x 0 1 − x

r1 + x 2 sin−1 1 Z 1 2 = 1 + √ dx. 2 2 0 1 − x ! d r1 + x 1 Observe that 2 sin−1 = √ , hence dx 2 1 − x2

r 1 + x 1 2 sin−1  r !2 Z 1 2 1 + x √ dx = 2 sin−1 2   0 1 − x 2 0 3π2 = . 8 So our original integral evaluates to

√ r1 + x 1 − x2 + sin−1 Z 1 2 3π2 √ dx = 1 + . 2 0 1 − x 16 √ • Easier Integral: Using the substitution u = 4x − 3,

Z 3 √ 1 Z 3 3 4x−3 dx = u · 3u du. 1 2 1 Now we can apply integration by parts:

1 Z 3 1 u · 3u 3 1 Z 3 u · 3u du = − 3u du 2 1 2 ln 3 1 2 ln 3 1

39 12 = − . ln 3 (ln 3)2

c UNSW Mathematics Society 2020 2020 MathSoc Integration Bee Finals Solutions • Question 1: First, notice that the integrand can be simpliﬁed into a product: √ √ 3 5 ∞ x x x Y 1 − 1 √ √ √ ··· = x( 2k−1 2k ). x 4 x 6 x k=1 The product can be moved into the exponent as a sum, giving us the expression

∞ Y 1 − 1 P∞ 1 − 1 x( 2k−1 2k ) = x k=1( 2k−1 2k ). k=1 The sum in the exponent is the taylor series of ln (1 + x) evaluated at x = 1, i.e. ln 2. So our integral becomes √ √ Z x 3 x 5 x Z √ √ √ ··· dx = xln 2 dx x 4 x 6 x x1+ln 2 = + C. 1 + ln 2

• Question 2: We simplify the expression by logarithm laws:

Z x 2x 2x Z x ex ln ex xx dx = ex x2x (1 + ln x) dx.

By using the substitution u = xx,

Z x Z ex x2x (1 + ln x) dx = ueu du

= (u − 1)eu + C x = (xx − 1)ex + C.

• Question 3: First, we must simplify the integrand:

∞ ∞ Z 1 Y x Z 1 Y x 1 − tan2 dx = 2 − sec2 dx x 2k x 2k k=1 k=1 ∞ Z 1 Y x x = sec2 2 cos2 − 1 dx x 2k 2k k=1 ∞ Z 1 Y x x = sec2 cos dx x 2k 2k−1 k=1 ∞ Z cos x Y x = sec dx. x 2k k=1 We can simplify the product in the integrand now, by using trig identities. x ∞ N 2 sin Y x Y 2k sec k = lim x 2 N→∞ sin k=1 k=1 2k−1 N x 2 sin N = lim 2 N→∞ sin x x = . sin x

c UNSW Mathematics Society 2020 Hence our integral becomes ∞ Z 1 Y x Z cos x 1 − tan2 dx = dx x 2k sin x k=1 which is integrable into a logarithm: Z cos x dx = ln (sin x) + C. sin x √ • Question 4: Using the substitution tan θ = ex − 1, ∞ π Z x Z 2 √ dx = −4 ln (cos θ) dθ. x 0 e − 1 0 π Substituting u = − θ, 2 π π Z 2 Z 2 ln (cos θ) dθ = ln (sin u) du. 0 0 We can add these two integrals together: π π Z 2 Z 2 2 ln (cos θ) dθ = ln (sin θ cos θ) dθ 0 0 π Z 2 = ln (sin 2θ) − ln 2 dθ 0 π Z 2 π ln 2 = ln (sin 2θ) dθ − . 0 2 For this resulting integral, we substitute φ = 2θ: π π Z 2 1 Z ln (sin 2θ) dθ = ln (sin φ) dφ 0 2 0 π Z 2 = ln (sin φ) dφ 0 π Z 2 = ln (cos φ) dφ. 0

Hence π Z 2 π ln 2 ln (cos θ) dθ = − . 0 2 So our integral evaluates to Z ∞ x √ dx = π ln 4. x 0 e − 1 • Question 5: We seek a reduction formula for this integral. Denote the original integral as I(m, n), then applying integration by parts we get Z 1 xm+1(1 − x)n 1 n Z 1 xm(1 − x)n dx = + xm+1(1 − x)n−1 dx 0 m + 1 0 m + 1 0

n I(m, n) = I(m + 1, n − 1). m + 1 I(m + n − k, k) Taking the product of the ratios from k = 1 to k = n: I(m + n + 1 − k, k − 1) n n Y I(m + n − k, k) Y k = I(m + n + 1 − k, k − 1) m + n + 1 − k k=1 k=1 I(m, n) m! · n! = . I(m + n, 0) (m + n)!

c UNSW Mathematics Society 2020 1 Evaluating I(m + n, 0) = : m + n + 1 Z 1 I(m + n, 0) = xm+n dx 0 1 = . m + n + 1 So our integral is m! · n! 1 I(m, n) = (m + n)! m + n + 1

m! · n! I(m, n) = . (m + n + 1)!

c UNSW Mathematics Society 2020