Maths ALevel C4
Rates of change and Differential Equations
… First order differential equations are of the form:
dy/dx = f(x)g(y)
so, for example:
dy/dx = 4xy²
They are often formed from a situation involving rates of change, such as water leaking out of a bucket.
To analyse a rates of change situation and form a differential equation:
1 – Write down a relationship between the variables using the geometry of the situation (you may need to use similar shapes to eliminate unwanted variables). Then differentiate this relationship. For an example involving an inverted cone of height 20cm and base radius 5cm filling up with water:
if V = (1/3)πr²h, and by similar shapes 20/5 = h/r → r = h/4
then we can eliminate r:
V = πh³/12
and so
dV/dh = πh²/4
2 – Write down the rate of change data given in the question, looking out for any units that are given. For example:
“the rate of water flowing into the cone is 6cm³/min”.
dV/dt = +6.
3 – Write down what differential you are looking for (remembering that ‘rate’ means d…/dt). For example:
“Find the rate of change of height of liquid.”
dh/dt = ?
4 – Use the chain rule to find the differential you’re looking for. Note that you may need to ‘flip’ one of the differentials (in this case we flip dV/dh):
dh/dt = dh/dV x dV/dt
dh/dt = 4/(πh²) x 6 = 24/(πh²)
… Translating the words into a differential equation. Look out for words such as:
“… at a rate of…” meaning “d(variable)/dt”
“… is proportional to…”, meaning “…=k(…”
“… is inversely proportional to…”, meaning, “…= k / …”
“… the product of…”
…any words meaning adding to, such as “enters”, “fills”, “raises”, “increases”
…any words meaning a NEGATIVE rate of change, such as “leaves”, “empties”, “decreases”
… When you are asked to solve a differential equation, rearrange the equation so variables of the same type are on each side. To do this, you will have to ‘separate the variable’ so that (for example) the dy/dt is separated into dy on one side and dt on the other.
Then integrate both sides remembering to include the constant +c on just one side. You will have found the ‘general solution’.
Now, sub in the conditions (e.g. “Initially the height is 20cm…” means t = 0, h = 20) to find the constant. You will then have found the ‘particular solution’.
… Sometimes you will have to form the differential equation by first finding two rate equations and then combining them, for example if you are asked to find an expression for dh/dt, the rate at which the height of water is dropping in a leaking cylinder:
“The rate at which water leaks out of the cylinder is proportional to the volume of water remaining”: so we can write:
dV/dt = – V
Now form an equation using geometry to link the volume and height of water in the cylinder: so:
V=Ah
where A is the cross sectional area of the cylinder.
Then differentiate to get:
dV/dh = A.
Then use the chain rule:
dh/dt = ( dV/dt )( dh/dV ) = (V)(A) = AV.
Note that you may then need to substitute V using equation 2, so:
dh/dt = A(Ah) = (A²)h
… Finding the general solution to a differential equation by ‘separating the variables’. The general solution is with the unknown constant of integration. Note that you can change the constant as you simplify the equation, for example:
if y/2 = e^x + A,
then
y = 2e^x + B as we have made the constant 2A = B
… The particular solution to a differential equation is when you use the conditions (e.g. x = 0 when y = 1) to find the value of the constant of integration.
… How would you make y the subject of:
e^y = A
… How could you simplify e^(3x+C) ?
… If there is no ‘y’ in the differential equation, imagine there is a ‘1’ in the integration, for example:
x + dy/dx = 5
dy/dx = 5x
dy = (5x)dx
∫1dy = ∫(5x)dx
and the next line would be…?
… ln(e) = ?
… Verifying a solution to a differential equation, e.g. If dy/dx = x(1x) then verify that x = 1/(1+e^t) is a possible solution.
To do this, we could either solve the differential equation (by splitting the variables and then integrating using partial fr actions), or we could more simply:
 Differentiate x = 1/(1+e^t) to find dy/dx.
 Then substitute x = 1/(1+e^t) into x(1x) and that should equal the dy/dx value calculated in step 1.
… Finding a differential equation that is the origin of a given general solution, and eliminating the arbitrary constant in the process, for example:
y² = Ax Could be our general solution, which differentiates to…
2y dy/dx = A
but A = y²/x, from the general solution, so
2y dy/dx = y²/x
dy/dx = y/2x , which is the differential equation.
If we then solved this differential equation (via integration), we would end up with the original equation:
y² = Ax
C4 Vectors:
… The unit vectors i, j and k are used to describe vectors, for example:
r = 3i – 4j + k
This can also be written as a vertical column vector in a single bracket:
3
4
1
Note that a column vector only gives us information about which direction and distance to travel. It can start from any position in space.
… Coordinates of a point are usually written in the horizontal form, e.g. (3 , 4 , 1). This is the position of the point in space, relative to an origin (0,0)
… Finding the vector between two points, e.g. AB = OB – OA
… The magnitude of a vector can be found by via Pythagoras using the coefficients of each unit vector. For example for the vector r = 3i – 4j + k, the magnitude of r is given by:
r² = (3)² + (4)² + (1)²
Calculating the magnitudes of a vector is often required when a question asks for the ‘distance’ or ‘length’ between two points.
… If you have a vector p, then a unit vector which is parallel to p is given by:
unit vector = p / p
where the sides of the triangle will be the i and j parts of the vector.
… To find the angle of a vector to the horizontal or vertical, try drawing a rightangled triangle with the vector as the hypotenuse. Then use:
tanθ = opp/adj
… To find the angle between a vector and the z coordinate axis (for example), first find the magnitude r of the vector r.
Then, use
cosθ = Adjacent / Hypotenuse
cosθ = (k vector component of vector) / r
… To find the ratio in which a line is intersected, or to prove that two lines intersect, you will first need to define multiplying constants λ and μ which take you some way along each line to the point of intersection.
For example, the vector OA is split by an intersecting line BC. The point of intersection is P.
Do draw the situation before diving into the vector algebra!…
B

O——P—————A

C
Then we can say:
OP = λOA
and
BP = μBC
If we needed to show that OP is the bisector of OA, then we need to show that λ=1/2 …
To do this, find another way to make vector OP by finding a different route that moves along each intersecting line. Use BP = μBC in this second vector for OP.
We then have two equations for OP. We can eliminate OP from these equations and then compare coefficients to form two simultaneous equations to find the values of λ and μ.
Instead of using λ and μ, we could form a vector for part of the line segments using a ratio, for example:
OP = [ 1/(1+k) ] a
(for example, if k=1, then OP=a/2 and we have a bisector).
… The ‘Vector equation of a straight line’ is the position vector of any point on the line. First find a position vector for a point ON the line (relative to the origin), then add a DIRECTION vector multiplied by a scaling parameter to ‘slide’ up or down the line. For example:
r = (2,1,3) + t(1,0,2).
Note that a vector equation of a straight line can be squashed into a single vector, for example in this case: (2+t , 1 , 32t) which is useful in vector calculations.
You can choose any direction vector which is parallel to the line. For example, if points P and Q lie on the line, then you could calculate vector PQ and use that as your direction vector.
… To prove that a point, e.g. (1, 0, 2) lies on a line, equate the components of the ‘squished’ form of the equation of the line with the point’s coordinates. For example:
( 2 – λ ) ( 1 )
( 0 ) = ( 0 )
( λ + 1 ) ( 2 )
In this case, λ = 1 works for all three x, y and z components, therefore the point lies on the line.
… To find the Cartesian equation of a line from its vector equation, squish the x, y and z components together and then list your three equations for x=… , y=… and z=…
Make the scaling parameter the subject for each equation and then eliminate the scaling parameter. You will usually get a ‘threesided equation’, for example:
(x1)/2 = y/3 = z – 1
… The ‘Scalar (dot) Product’ is the multiplication of the parallel components of two vectors. (For example, this is used to calculate the work done by a force through a distance in physics).
… The Scalar Product formula (sometimes known as the dot product) can be used to find the angle between two vectors:
a.b = abcosA
Note that vectors a and b are usually ‘direction type’ vectors such as AB and AC (which would find us the angle between lines AB and AC)
… The scalar product of two vectors = sum of the products of the respective i, j and k coefficients. For example, if a = 3i – 4j + k, and b = 2i + k, then
a.b = (3)(2) + (4)(0) + (1)(1) = 7
… You can use the scalar product to find points on a line to satisfy a perpendicular criterion where a.b = 0
… When finding the position vector of a point on a line, you will usually need to start by defining that position vector in terms of the vector equation of the line.
So, if we know that a point A lies on a line with the equation
r = (2,1,3) + t(1,0,2)
then we can start by writing down:
The position vector ra = (2,1,3) + t(1,0,2)
= (2+t , 1 , 32t) in its squashed form 🙂
… To find the shortest distance between a line and a point P,
 Find the vector AP where A is the position vector of a point on the line in terms of a scaling parameter (i.e. the vector equation of the line).
 Use the scalar product result for two perpendicular lines: AP.b = 0 where vector b is the direction vector of the straight line. This will give you the value of the scaling parameter.
 Substitute the scaling parameter into AP and use Pythagoras to find the length of the vector (the modulus).
… If point A lies on a line and OA is perpendicular to the line, then to find the coordinates of A:
OA can be written as the vector equation of the line in its ‘squashed’ form, in terms of the scaling parameter, t.
Use OA.(direction vector of the line) = 0 to find the value of t.
Substitute t into the vector equation of the line to find the coordinates of A.
… Finding the Intersection coordinates of vector lines: equate each component to form three equations in terms of the parameters. Often one of the equations will only have one unknown. Use the third equation as a ‘check’ once you have found the values of the parameters from the previous two equations.
Then substitute one of the parameters into one of the lines to find the coordinates of intersection.
… When two vectors diverge, the scalar product formula will give you the angle between them. This could be acute (giving a positive value for cosA) or obtuse (giving a negative value for cosA).
If you come up with an obtuse angle, simply subtract it from 180° to find the acute angle.
… A direction vector can be any length, and even the opposite direction. So the vectors:
(2 , 1 , 3)
(4 , 2 , 6)
(2 , 1 , 3)
Could all be used as the direction vector of the same line!
… A unit vector always has a magnitude of 1. To find a unit vector, divide each component of a vector by the vector’s total magnitude:
If r = 0i – 4j + 3k
Magnitude = (0 + 16 + 9)^(0.5) = 5
So the unit vector which points in the same direction as r would be:
(0/5)i – (4/5)j + (3/5)k
= – (4/5)j + (3/5)k
… Lines are ‘Skew’ if they do NOT intersect and are NOT parallel.
… Parallel vectors: one will be a multiple of the other. For example:
AB = (2, 5, 3) in column vector form
and
CD = (4, 10, 6)
are parallel because CD = 2AB
–
The equation of a plane
Method 1
… If we know three different points A, B and C which lie on a plane, then we can define that plane.
The first step is to get ‘on’ the plane by choosing the position vector of one of the points, say OA.
Then we can travel within the plane some distance λ in the direction AB, and some distance μ in the direction AC. Different values of λ and μ will move us to any position on the plane, so:
r = OA + λAB + μAC
Method 2
… The other way of defining a plane is if we know a vector which is perpendicular to the plane. This is called the normal vector, n, which in column vector form would be:
[n1 , n2 , n3]
We will also need a known point A which lies on the plane. Then if a general point R also lies on the plane with a position column vector [x , y , z]
Then
AR.n = 0
This can be written
(ra).n = 0
r.n – a.n = 0
n1.x + n2.y + n3.z – a.n = 0
We can call a.n = d which is a constant scalar. This fixes the plane in space. Different values of d will result in lots of parallel planes hanging in space!
n1.x + n2.y + n3.z + d = 0
This is now a Cartesian form for the equation of a plane. Notice how the coefficients of x, y and z give the normal vector to the plane.
–
… To find the Cartesian equation of a plane from its vector equation, squish the x, y and z components together and then list your three equations for x=… , y=… and z=…
Then try to combine these three equations to eliminate the scaling parameters to form an equation in x, y and z.
Note that you may end up with an equation in only 1 or two variables, for example x = y – 2. This means that the value of z has no effect on x or y and so can be ignored (just like in 2 dimensions when the line x=2 doesn’t have a y part to it)
… To find the Cartesian equation of a plane from the normal vector and a point on the plane, use the normal coefficients. For example:
if n = 3i + 4j +5k
Then
3x + 4y +5z + d = 0
Then sub in the coordinates of a point in the plane and find the value of d.
–
Finding an intersection between a line and a plane
… For example, if a plane has an equation x + 2y – 4z = 3 and a line has equation r = [1,0,2] + λ[3,2,1]
Then we can ‘squash’ the equation of the line into the column vector:
[1+3λ , 2λ , 2+λ]
These are the x, y and z coordinates of a general point which lies on the line.
Now substitute these coordinates into the equation of the plane:
(1+3λ) + 2(2λ) – 4(2+λ) = 3
Which we can solve to find the value of λ, and hence the position vector of the point on the line which also lies in the plane, (i.e. the coordinates of intersection).
–
Finding the angle between a plane r.n = k and a line r = a + λb
… Here, b is the direction vector of the line
… Vector b and the normal n create a rightangled triangle with the adjacent side on the plane.
… The angle A between the normal and the direction vector can be found using the scalar product:
b.n = bncosA
… However, we need the angle, B, between the direction vector and the adjacent side of the triangle (in the plane).
B = 90 – A
because the adjacent and the normal are at right angles to each other.
This means that A = B + 90
So
cos(A) = b.n/(bn)
cos(90B) = b.n/(bn)
sin(B) = b.n/(bn) because cos(90B) = sin(B) using transformations.
–
Finding the shortest distance from a point B (1,0,1) , to a given plane.
… If you know the Cartesian equation of the plane, for example: 2x – y + 3z = 1, then we know the normal vector to the plane, n, which is:
[2 , 1 , 3] in column vector form
Then construct a vector equation of the line perpendicular (normal) to the plane.
r = OB + λ(n)
r = [1,0,1] + λ[2,1,3]
Now find the intersection of this line with the plane (see above) to work out the value of λ, and so we are then able to find:
1) the intersection position vector of point A,
2) the vector AB, and
3) AB using Pythagoras
–
Proving that a vector n is perpendicular to a plane which contains three points, e.g. A, B and C.
Use the scalar (dot) product result that:
a.b = 0
for when vectors a and b are perpendicular.
a can be our ‘n’ vector
b can then be either vector AB or AC (or some other combination of A, B and C)
Test that
n.AB = 0
and
n.AC = 0
If so, then AB and AC are both perpendicular to the vector n, and so n is perpendicular to the plane (because A, B and C) lie on the plane.
Partial fractions
… The denominators of the partial fractions must have a Lowest Common Multiple which is the same as the denominator of the original fraction. Consider all possible factors of the original denominator to form your partial fractions. For example:
2/(x1)(x+2)² = A/(x1) + B/(x+2) + C/(x+2)²
Note that in this example we have a repeated factor of (x+2)² and so have used both (x+2)² and (x+2) as denominators in the partial fractions.
… If the power of x in the numerator is the same or higher than the power of x in the denominator, you have an improper fraction. You’ll first need to do a long division to split up the improper fraction into a constant + proper fraction. Then go into partial fractions.
The alternative to using long division is:
 if the numerator is the SAME order, then add a constant, e.g. ‘A’
 If the numerator is one order higher, then add the terms A + Bx
… Once you have used long division on an improper fraction, then you may get a remainder, which will still need to be divided by the divisor. For example:
(x²+4)/(x²+2x3) = 1 + (2x+7)/(x²+2x3)
numerator/ divisor = quotient + remainder/divisor
We can then use the usual partial fractions method on (2x+7)/(x²+2x3):
(2x+7)/(x²+2x3)
(2x+7)/(x+3)(x1) = A/(x+3) + B/(x1)
etc.
… If you have a quadratic factor in the denominator, for example (x² + 1) then you will need to use (Ax + B) in the numerator of the partial fraction.
This is because (x² + 1) can actually be factorised further using complex numbers, but this is beyond the scope of C4 maths!
… During long division of algebra, remember Dad ÷, Mum ×, Sis , Bro (bring down). You may also need to add in other lower powers of x inside the ‘bus stop’, for example x³ + 0(x²) + 0x + 0
… Finding the coefficients e.g. A, B and C by choosing values of x to eliminate terms (including x = 0). Once you have exhausted the values of x that you can choose, either use x = 0, or consider the coefficients of x or x² on the LHS compared to the RHS.
… Integrating by first using partial fractions to simplify the terms in the integral. The integrals often produce ln functions via the rule that ∫f'(x)/f(x)dx = lnf(x) + C.
… Notice that modulus signs are used with the ln function produced from:
∫f'(x)/f(x)dx = lnf(x) + C
The reason why the modulus sign is needed is covered in more detail at:
http://mathhelpboards.com/calculus10/doyouneedabsolutevaluearoundargumentlogln7032.html
There is no need to use the modulus signs with ‘normal’ logarithms as log(x) is not defined for x ≤ 0.
… Partial fractions can written in ascending powers of x using binomial expansion. Use the second formula in the formulae booklet because the power (n) will be negative.
Binomial Expansion (when the power n < 0 or a fraction)
… When the power is a negative and/or a fraction, the expansion of (1 + x)^n will create an infinite series of expansion terms.
… For the binomial expansion of (1 + x)^n, x < 1 is the condition necessary for the expansion “to be valid”, which means that using the first three or four terms will give a good approximation to the true value (each successive term in the series will get smaller)
For example, if (1 + 2x)^0.5
then the expansion will be valid if 2x < 1
so x < 1/2
… You may need to rewrite a binomial in order to get it into the form:
(1 + x)^n
For example:
(2 + x)^1
= [2(1 + x/2)]^1
= (2^1) (1 + x/2)^1
= (1/2) (1 + x/2)^1
Now we can proceed to expand this using our “x” as being x/2.
… Note that the previous expansion that you learnt in C2 for (a + b)^n is only useful for when the power, n, is a whole number greater than zero. This does not create an infinite series.
(The new C4 expansion method for (1 + x)^n will also work for whole number n values, but it is much more work!)
… When expanding a binomial, use brackets (..) for substitutingin the x terms so you can see the effect of squares or cubes on the x coefficient.
… Work through each line of simplification a step at a time – keep checking your coefficients as it is very easy to make mistakes in binomial expansions!
… The terms of a binomial expansion will usually have alternating or constant signs. So if you see that your terms have signs, for example + … – … – … Then you should check your working for an error in the signs somewhere!
… Using a binomial expansion to find an approximation to (for example) a root of a nonsquare number. E.g. (3.999). First substitute the suggest value of x into the original expression and simplify. You may find that the simplification is just a small modification away from the approximation they are asking for.
However, sometimes you may need to work out a value of x which also satisfies the condition needed for the expansion to converge (1 + x)^n, x < 1. For example
(8 – 9x)^(1/3)
[8(1 – 9x/8)]^(1/3)
2(1 – 9x/8)^(1/3) so this expansion is valid if x < 8/9
“Find an approximation for (7100)^(1/3)”
The trick here is to take out a factor of a simple cube number such as 1000, so we get
10(7.1)^(1/3)
Now 8 – 9x = 7.1 making x = 0.1, which small enough to satisfy x < 8/9
x = 0.1 can then be substituted into the original expansion.
You may often need to treat each term of the expansion separately when you substitute the value of x back in in this way you will be able to see the number of decimal places that each term contributes.
You may need to add the terms up manually (or with a graphics calculator) to give the number of decimal places the question requires!
Integration by parts
Like integration by substit ution, integration by parts also enables us to rewrite the integral so that it becomes easier to integrate.
… You’ll need this method when finding the integral of two multiplying functions (product rule in reverse).
… ∫(uv’ dx) = uv – ∫(u’v dx)
… Always try to choose u so that du/dx (or u’) is simpler.
… Integrals involving ‘ln’ usually involve making u = the ln term.
… The standard ‘trick’ for integrating ln(x) is to make u = ln(x) and dv/dx = 1
… Sometimes you will have to use integration by parts TWICE in order to form a simpler integral of u’v dx.
… Occasionally you will see ‘cycling’ where the integral of u’v dx cycles from one expression back and forth to another. In this case, spot the cycle where the integral is the same as the original and then add it to both sides to eliminate from the RHS.
Implicit differentiation
… To find the differential of a ‘y’ term with respect to x, first differentiate it with respect to y and then multiply it by dy/dx (this is a result of the chain rule).
For example,
y² + x³ = 4
If u = y² then du/dx = du/dy x dy/dx = 2y(dy/dx), and so
2y(dy/dx) + 3x² = 0
dy/dx = 3x²/2y
… To find stationary points, set dy/dx = 0 as usual. However, sometimes this may lead to another equation such as:
2x + y = 0
To find the coordinates of the stationary points, solve this new equation simultaneously with the original equation of the curve.
… Using implicit differentiation with the product rule.
… Here is a worksheet for extra practice: http://cdn.kutasoftware.com/Worksheets/Calc/03%20%20Implicit%20Differentiation.pdf
… You’ll also be using implicit differentiation with the chain rule and possibly the quotient rule :).
… When finding the gradient of a curve (or finding the value of d²y/dx² to identify the nature of a stationary point), it is often MUCH easier to substitute the coordinates immediately once you have implicitly differentiated (rather than try to simplify and factorise to get dy/dx as the subject).
… Rewriting a given equation into a more recognisable form can be a useful tool for awkward implicit differentiation questions. For example, if:
tanθ = 6y / (160 + y²)
Then we could rewrite this as:
tany = 6x / (160 + x²)
It’s then much easier to ‘see’ the usual implicit differentiation approach, so we get:
sec²y dy/dx = (use the quotient rule for the RHS…)
Parametric Equations
… Parametric equations help us to express a complicated function in a simple way using x, y and a ‘linking’ parameter such as ‘t’.
… To find the Cartesian equation, substitute one equation into the other to eliminate ‘t’.
However, sometimes the substitution method won’t work… In these cases, try + – ÷ or × the equations together.
Often you’ll find that a key term will cancel so that you can find t in terms of x and y. Then go ahead and substitute as usual.
For example,
x = t² + t
y = t² – t
Here, try subtracting: x – y and making t the subject. Then substitute t into either equation!
… You may also need to square both equations and add or subtract them, especially for equations containing trigonometry terms (use identities to simplify the result).
… The allowed set of values for the input parameter, t, can be thought of as the ‘domain’ of the parametric function.
There are then 2 ranges; one for the possible set of values for x, and the other for the possible set of values for y.
Once you have found the range of possible x values (by sketching the graph of x vs t), you can then treat this as the domain for the cartesian equation.
… Parametric functions can be differentiated by using the chain rule. For example:
y = Asin(t) x = 3t – cos(t)
dy/dt = Acos(t) dx/dt = 3 + sin(t)
dy/dx = dy/dt × dt/dx = Acos(t) / (3 + sin(t))
… You may need to substitute the parametric equations for x and y into the equation of straight line in order to solve for the parameter. For example:
“A curve has parametric equations x = 2cos(t) and y = 3sin(t). Find the coordinates of the points on the curve for which the tangent to the curve cuts the x axis at (4,0)”
First find dy/dx as this will be our gradient, m value.
Then use the equation y – y0 = m(x – x0) substituting in the coordinates (0,4) for the (x0, y0) values and x = 2cos(t) and y = 3sin(t). You’ll form a quadratic equation which you should be able to solve for t.
… Parametric questions often involve using the gradient dy/dx in terms of t. To find stationary (maximum or minimum) points use dy/dx = 0.
To find points at which the gradient is vertical (i.e. infinity!) set the denominator of the dy/dx expression to equal zero.
For example,
if dy/dx = (sin(t) + 1) / (sin(t)cos(t))
Then a vertical gradient is when sin(t)cos(t) = 0
… Integrating parametric equations by substitution (e.g. to find the area under a curve):
∫ y dx =∫ y dx/dt dt
If x = 4cos(3t) and y = 4sin(t), then to find Integral [ y dx ], make the substitutions:
y = 4sin(t)
and
dx/dt = 12sin(3t) so dx = 12sin(3t)dt
∫ y dx =∫ 4sin(t)(12sin(3t)) dt
∫ y dx =∫ 48sin(t)sin(3t) dt
… Remembering to change the limits using t = arcsin(x/4)
Note that if you get more than one answer for t, that you should check which one makes sense by substituting that value into the parametric for y.
This can then be integrated by using a double angle formula (or by integrating by parts twice!) to find the area under the curve.
… You can plot a parametric equation using a table of t, x and y values. Choose several values for t and find the corresponding x and y values.
The ‘table’ mode on your calculator is useful for doing this (although you’ll have to find the x and y values separately).
Integration by Substitution
Integration by substitution enables us to rewrite the integral so that it becomes is easier to integrate.
… Exam questions will usually suggest a substitution to use. If not, then try an obvious looking bracket.
For example to integrate x/(x² + 3) you could use the substitution u = x² + 3
… Differentiate the substitution to find a substitution for dx.
… Substitute for u and du first and see if there is any simplification possible (usually between numerator and denominator) before substituting for the last ‘x’ terms.
… You may need to use trigonometry identities to simplify a given substitution.
… Remember to change the limits into the new ‘u’ variable (alternatively substitute the ‘u’ terms in the square limit brackets back into the original ‘x’ terms, so then you won’t need to change the limits).
… Take coefficients out of the integral where possible to simplify working.
… The following substitutions are recommended for the more difficult cases, however, a substitution will usually be suggested in the exam:
(ax + b)^n … u = ax + b
√(a – bx²) … x = √(a/b) sinu
√(a + bx) … x = √(a/b) tanu
√(bx² – a) … x = √(a/b) secu
e^x … u = e^x
ln(ax + b) … e^u = ax + b
Reverse Chain Rule (“by inspection”) – or “guess – check – modify”
… Integrals such as
∫e^(3x) dx,
∫sinA(cosA)^3 dx,
and
∫x(x² – 3)^4 dx
… can often be ‘guessed’ by trying a likely result and differentiating it to see if you get close to the original function within the integral.
In the example ∫sinA(cosA)^3 dx, you could try differentiating (cosA)^4 and then modifying by a factor as required.
Do refer to the list of standard integrations and differentiations in your formulae booklet and identify any that might ‘fit’.
… The standard result for reciprocal type Integrals:
∫f'(x)/f(x) dx = lnf(x) + c
Note that you may need to modify the numerator by a numerical factor (with an ‘antifactor’ outside the integral!) so that it becomes the differential of the denominator. So for example:
∫x/(x² + 3) dx
= (1/2) ∫ 2x/(x² + 3) dx
= ln (x² + 3)  + c … Because 2x is the differential of (x² + 3)
Other useful integrals
… ∫sin³x dx can be found by using the identity sin²x = 1 – cos²x:
= ∫sinx(1 – cos²x) dx
= sinx – sinxcos²x dx
Then use Guess – Check – Modify (reverse chain rule) to integrate sinxcos²x (guess cos³x to start)
… The integral of (sinA)(cos²A) = (1/3)cos³A via GuessCheckModify
… The integral of (cosA)(sin²A) = (1/3)sin³A via GuessCheckModify
… The integral of tan²A can be found by using the identities
tanA = sinA/cosA
and
sin²A + cos²A = 1
Then use the standard result (from your formulae booklet) that
∫sec²x dx = tanx + c
… The integral of sin²A and cos²A can be worked out by first substituting the squared trig function with a double angle trig function:
cos(2A) = cos²A – sin²A
cos(2A) = 2cos²A – 1
So cos²A = (cos(2A) + 1) / 2 … Which is much easier to integrate 🙂
… The compound angle formulae for sin(A±B) can be combined to give the result:
½[sin(A+B) + sin(AB)] = sin(A)cos(B)
This can be used to split up integrals such as ∫sin(5x)cos(2x) dx :
= ∫½[sin(5x+2x) + sin(5x2x)] dx
= ½∫sin(7x) + sin(3x)] dx