Integration (5)

Question 1

Use the substitution u = x^2 - 2 to find ∫ 4x(x^2 - 2)^4 dx

Answer 1

As u = x^2 - 2
du/dx = 2x
dx = (1/2x) du

∫ 4x(x^2 - 2)^4 dx
= ∫ 4x(u)^4 (1/2x) du 
= ∫ 2u^4 du
= (2/5) u^5 + C
= (2/5) (x^2 - 2)^5 + C

Question 2

Use the substitution u = tan x to find ∫ (sec^4 x)/¬(tan x) dx

Answer 2

As u = tan x
du/dx = sec^2 x
dx = (1/sec^2 x) du

As sec^2 x = 1 + tan^2x
sec^2x = 1 + u^2

∫ (sec^4 x)/¬(tan x) = ∫ ((sec^4 x)/¬(tan x)) * (1/sec^2 x) du
= ∫ (1 + u^2 / ¬u) du

= ∫ ((1/¬u) + ((u^2)/¬u)) du
= ∫ (u^-(1/2) + u^(3/2)) du
= 2u^(1/2) + (2/5)u^(5/2) + C = 2¬(tan x) + (2/5)¬(tan^5 x) + C

Question 3

Find the value of 2∫1 (8/x)(lnx + 2)^3 dx using the substitution u = ln x. Give answer to 4 dp

Answer 3

Using u = lnx
du/dx = 1/x
dx = x du

Change the limits
When x = 1, u = ln 1 = 0
When x = 2, u = ln 2

Substitute into the integral
2∫1 (8/x)(lnx + 2)^3 dx
Becomes ln2∫0 (8/x)(u + 2)^3 x du
= ln2∫0 8(u + 2)^3 du

[2(ln2 + 2)^4] - [2(0 + 2)^4]
= 105.213… - 32 = 73.21 (4 s.f.)

Question 4

Use a suitable substitution to show that ∫(1/((¬u)(¬u - 1)^2)) du = (-2/((¬u) - 1)) + C

Answer 4

Let v = ¬u - 1
dv/du = 1/(2¬u)
du = 2¬u dv

∫(1/(¬u)(v^2)) 2¬u dv = ∫ 2v^-2 dv
= 2v^-1 + C = (-2/(¬u - 1)) + C

Question 5

pi∫pi/2 x sin x dx

Answer 5

u = x
dv/dx = sin x

du/dx = 1
v = -cos x

= pi[- x cos x]pi/2 - pi∫pi/2 (- cos x dx)
= pi[- x cos x]pi/2 + pi[sin x]pi/2

= [(-pi cos pi) - ((-pi/2) cos (pi/2))] + [(sin pi) + (sin pi/2)]
= [pi - 0] + [0 - 1] = pi - 1

Question 6

Integrate the function f(x) = -sin x (3 cos^2 x) with respect to x

Answer 6

if f(x) = cos x, f'(x) = -sin x and n=2
So (n + 1)f'(x)[f(x)]^n = 3 * -sin x * cos^2 x = -sin x(3cos^2 x)
∫ -sin x(3cos^2 x) = cos^3 x + C

Question 7

∫ (12x^3)(2x^4 - 5)^5

Answer 7

f(x) = 2x^4 - 5
f’(x) = 8x^3
n = 2
2 + 1 = 3

∫ 3(8x^3)(2x^4 - 5)^5 dx
∫ (24x^3)(2x^4 - 5)^5 dx = (2x^4 - 5) + C

Divide by two to match the original integral
∫ (12x^3)(2x^4 - 5)^5 = (1/2)(2x^4 - 5) + C

Question 8

Find the particular solution of dy/dx = 2y(1+x)^2 when x=-1 and y=4

Answer 8

f(x) = 2(1+x)^2 AND g(y) = y
Rearranging the equation gives: (1/y) dy = 2(1+x)^2 dx

∫ (1/y) dy = ∫ 2(1+x)^2 dx
ln|y| = (2/3)(1 + x)^3 + C

ln|4| = (2/3)(1 + -1)^3 + C
ln4 = C

The solution is ln|y| = (2/3)(1 + x)^3 + ln4

Question 9

Find the general solution to the differential equation (dy/dx) = (cosx cos^2y)/(sinx), 0 < x < pi, -(pi/2) < y < (pi/2)

Given that y=0 when x=pi/6, solve the differential equation above

Answer 9

(1/(cos^2 y)) dy = (cos x/sin x) dx
∫ (1/(cos^2 y)) dy = ∫(cos x/sin x) dx
∫ sec^2 y dy = ∫cot x dx
tan y = ln|sin x| + C

tan 0 = ln|sin (pi/6)| + C
0 = ln|(1/2)| + C
-ln2 + C
C = ln2

So tan y = ln|sin x| + ln2

Question 10

A company sets up an advertising campaign to increase sales of margarine. After the campaign, the number of tubs of margarine sold each week, m, increases over time, t (in weeks), at a rate that is directly proportional to the square root of the number of tubs sold

Formulate a differential equation in terms of t, m and a constant k

At the start of the campaign, the company was selling 900 tubs of margarine per week. Use this to solve the differential equation, giving m in terms of k and t

Hence calculate the number of tubs sold in the fifth week of the campaign, given that k=2

Answer 10

(dm/dt) = k¬m, k>0

(dm/dt) = k¬m
(1/¬m) dm = k dt
∫ m^(-1/2) dm = ∫ k dt
2m^(1/2) = kt + C
m = ((1/2)(kt + C))^2
m = ((1/4)(kt + C))^2

At the start of the campaign, t=0. Putting t=0 and m=900 into the equation gives:
900 = ((1/4)(0 + C)^2
3600 = C^2
C = +-60

When t=0, (dm/dt) = (1/2)(kt + kC) = (1/2)(kC)
Since k>0 and (dm/dt)>0 as sales are increasing, C must be positive at C=60.
Gives the equation m = ((1/4)(kt + 60))^2

Substituting t=5 and k=2 into the equation gives
((1/4)(10 + 60))^2 = 1225 tubs sold