Integration (6) Flashcards
What to do if integration is a standard result?
Does it need scaling
What to think of when manipulating it to a standard result?
Think trig identities and expanding brackets
When can I use the reverse chain rule?
Is the numerator a derivative of the denominator
Is one factor related to the derivative of another
When can I split the numerator?
Is there a single term in the denominator?
When can I do partial fractions?
Does the denominator factorise?
When can I do algebraic division?
Is the fraction improper?
Use reverse chain rule for ∫cosxsin^2x dx
Consider: (sinx)^3
Differentiate: 3cosxsin^2x
Scaling: We want 1/3
(1/3)sin^3x + C
Use reverse chain rule for ∫x(x^2+5x)^3
Consider: (x^2+5x)^4
Differentiate: 8x(x^2+5x)^3
Scaling we want: (1/8)
(1/8)(x^2+5x)^4 + C
Use reverse chain rule for ∫(2x)/(x^2+1)
Consider: ln|x^2 + 1|
Differentiate: (2x)/(x^2+1)
Scaling we want: None required
ln|x^2 + 1| + C
Use u=2x+5 to find ∫x¬(2x+5) dx
Find x:
u = 2x+5
x = (u-5)/2
Find dx:
u = 2x+5
du/dx = 2
dx = (1/2)du
Substitute these into expression: ∫((u-5)/2) * u^(1/2) * (1/2)du ∫((u-5) * u^(1/2))/4 du (1/4) ∫(u-5) * u^(1/2) du (1/4) ∫u^(3/2) - 5u^(1/2)
Integrate simplified expression:
(1/4) (2/5)u^(5/2) - (10/3)u^(3/2) + C
(1/10)u^(5/2) - (5/6)u^(3/2) + C
Write answer in terns of x:
Replace u with (2x+5)
Calculate (pi/2)∫0 (cosx)¬(1+sinx) dx
u = 1 + sinx du/dx = cosx dx = (1/cosx) du
Sub these values into: u = 1 + sinx
x = pi/2, u = 2
x = 0, u = 1
2∫1 cosx * u^(1/2) * (1/cosx) du
2∫1 u^(1/2) du
= 2[(2/3)u^(3/2)]1
= (2/3)2^(3/2) - (2/3)1^(3/2)
Find the general solution to (1+x^2)(dy/dx) = xtany
(1+x^2)(dy/dx) = xtany (1/tany)(dy/dx) = x/(1+x^2)
∫ (1/tany) dy = ∫ x/(1+x^2) dx
∫ coty dy = ∫ x/(1+x^2) dx
ln|siny| = (1/2)ln|1+x^2| + lnk ln|siny| = lnk¬(1+x^2) siny = k¬(1+x^2) y = arcsin(k¬(1+x^2))
The rate of increase of a rabbit population (with population P, where time is t) is proportional to the current population. Form a differential equation and find its general solution
dp/dt = kP
(1/P) dp/dt = k
∫(1/P) dP = ∫k dt
ln|P| = kt + C
P = e^(kt + c)
P = Ae^(kt)
t=0, p=300
300 = Ae^0
A = 300
Determine the area bound between the curve with parametric equations x = t^2 and y = t + 1, the x-axis, and the lines x = 0 and x = 3
∫y (dx/dt) dt
x = t^2 dx/dt = 2t y = t + 1
x = 3, t = ¬3 x = 0, t = 0
¬3∫0 (t + 1)2t dt ¬3∫0 (2t^2 + 2t) dt = ¬3[(2/3)t^3 + t^2]0 = (2/3)*(¬3)^3 + (¬3)^2 = 2¬3 + 3