- We can think of integration in 2 different ways:
- As the inverse of differentiation;
- As a method to calculate area or volume by combining smaller parts
- We have already learned various integration techniques, including:
- Integration by parts; and
- Integration by substitution.
Let’s remind ourselves of integration by substitution by looking at these examples which require substitution of trigonometric identities. With definite integrals, take care to also change the limits.
- Integrate with respect to x
- Find the exact value of the integral
hyperbolic functions can often be very effective substitutions. For instance, we could calculate by substituting x=sec(u) (as tan2u=sec2u – 1 and (sec(u))’ = sec(u)tan(u)), but it is easier if we substitute x = cosh(u).
- As cosh2u – sinh2u = 1 and dx/du = sinh(u), so
- cosh2u – 1 = sinh2u; and
- dx = sinh(u)du.
So our integral becomes = u + c = cosh-1x + c
- Using appropriate substitutions (which may be hyperbolic or trigonometric, determine the result of:
If we have a quadratic expression on the denominator, sometimes trigonometric or hyperbolic substitutions can still be helpful, but we should first complete the square in the quadratic before substituting for the squared part:
- Integrate with respect to x:
Integrating Hyperbolic Functions
From what we previously saw differentiating hyperbolic functions, it should be clear that and
To integrate tanh x, we see that and notice that this is of the form , so integrates to give ln|coshx| + c. Similarly, .
- Find a solution for the following integrals:
Now let’s integrate inverse hyperbolic functions:
- Let u = cosh-1x and v’ = 1, so and v = x, so
- = =
Integrate tanh-1x with respect to x.
To introduce this concept, let’s consider integrating cos2x using integration by parts:
- = . Let u = cosx and v’ = cos x, so
- , so:
- , which we can rearrange to give:
- , or
Given that I3 = , find the result for I3 using integration by parts.
Now we will try to calculate
- . Let u=cosn-1x and v’=cosx, so
- = . Using sin2x=1-cos2x, we get:
- = ,
This rearranges to give the reduction formula valid for n≥2:
Below is an example of how we would use this to calculate for instance, which can be called I4.
If , show, using integration by parts, that for n≥2
We can use the same approach with hyperbolic functions, as shown below for cosh(2x):
- . Let u=coshn-12x and v’=cosh2x, so
- = . We use sinh22x = cosh22x – 1 to give:
- = . So
- , which rearranges to give:
- , which is valid for all n≥2
If , show that the reduction formula for n≥2 is .
Reduction formulae can also be used with functions that are neither trigonometric nor hyperbolic.
- Let’s use this technique to get reduction formulae for:
- . (note that here, once we have used integration by parts, we will need to rewrite x8 as x2(1+x6-1)).
If , show using integration by parts, that for n≥2.
For definite integration, we should use the reduction formula before substituting the limits.
Show that has reduction formula for n≥2 and hence find the exact value of
Arc Lengths and Surface Areas
By using the Leibniz notation we can get an idea of how to calculate an arc length.
In the below diagram of the curve we call a small section of the curve 𝛿s, at which point we can approximate it as a line segment with length (𝛿s)2 = (𝛿x)2 + (𝛿y)2.
If we divide all terms in the above equation by (𝛿x)2, we get: , so .
Taking the square root of both sides gives: .
It is by taking the integral at the limit as 𝛿x tends to zero that we thus get the cartesian equation for arc length:
Find the length of the curve y=coshx from x=0 to x=ln2
Curves given in parametric form
Here we still have that (𝛿s)2 = (𝛿x)2 + (𝛿y)2, but as our curve is given in terms of parameter t, we divide not by (𝛿x)2 but by (𝛿t)2 , to give: .
So following the same steps as before, from this we derive the equation for the length of an arc in parametric form:
If x = cos2t and y = sin2t, determine the length of the parametric curve from t=0 to t=π/4.
Curves given in polar form
Consider the diagram below (of curve r=e𝜽):
Be thinking of OPR as a sector, we can approximate RP (as 𝛿𝜽 -> 0) with r𝛿𝜽. Also, angle QRP tends towards a right angle. So at the limit, when 𝛿s can be approximated by a line, we can say that (𝛿s)2=(𝛿r)2 + (r𝛿𝜽)2, which gives us , which leads to our formula for the length of an arc in polar form,
If r = sin𝜽, determine the length of the polar curve over the interval [0,π/4]
Surface Area of Rotation
We can determine the surface area when the arc of a curve is rotated about an axis, as illustrated below:
If S is the surface area when the curve is rotated around the x-axis, then the small disc of surface area generated by a small arc 𝛿s is approximately 𝛿S = 2πy𝛿s.
Using similar arguments to these above, we can thus derive the following two results:
- For cartesian curves, ; and
- For parametric curves, .
Find the surface area when the curve is rotated around the x-axis from x=0 to x=1.
Givent that , find the surface area generated when the curve is rotated around the x-axis from x=2 to x=12.
The equation for the movement of a satellite is given by x = acos3t, y = asin3t, where a is constant. Determine the surface area formed when the curve is rotated around the x-axis from t=0 to t=π/2.
Rotating around the y-axis
If we rotated around the y=axis, the radius x appears in our formula rather than the radius y, so we have the following two results:
- For cartesian curves, ; and
- For parametric curves, .
Find the surface area generated by the curve y=x2 as it rotates about the y-axis between x=1 and x=2.
A curve is defined parametrically as x=cos22t, y=sin22t for 0 ≤ t ≤ π/4. Determine the surface area generated when this curve is rotated about the y-axis.
Limits of Areas
Let’s think about infinite series, e.g. .If we think of each term in the sequence as an infinitely thin rectangle of height f(n) we can then consider the sum of these areas as . This method can help us identify whether a series converges or diverges.
Consider for instance, the harmonic series, . We can show that this diverges:
= [lnx]n1, = ln(n). Clearly as n->∞, this tends to infinity, so the area under the curve y=1/x is infinite. Below we look at the curve alongside the series shown as set of bars:
Clearly the area of the bars is greater than the area under the curve, so as the area under the curve is infinite, so the area of the rectangles that correspond to the series is also infinite, indicating that the series diverges. We call this the integral test.
Consider . On the two graphs below we can see on the left graph that if we use the initial value of each rectangle to determine the rectangle’s height, then the area of the rectangles is greater than the area under the curve, but if we use the final value of each rectangle to determine the rectangle’s height, then the area of the rectangles is less than the area under the curve.
So the curve is an upper bound for , so to the area under the curve plus 1 is an upper bound for .
In summary, , and as n->∞ this upper boundary tends towards 2.
We can also use the first graph to identify a lower bound:
, so the lower limit as n-> is 1.
(a) Find the upper and lower bounds for
(b) Evaluate the infinite sum . If the series converges, give an estimate for the upper bound.
Let’s summarise the possible outcomes, where a curve is denoted by R and the area of a summation of rectangles is denoted by S:
- If R > S and R converges, then S also converges;
- If R < S and R diverges, then S also diverges;
- If R > S and R diverges, this does not imply anything about S;
- If R < S and R converges, this does not imply anything about S.
In the last two cases, we would need to reformulate the problem in order to tackle it.
Find the upper bound for and hence show that diverges.
Convergence and divergence can also be identified by comparing summations with known intervals. What would be a suitable comparison for the previous example?
Use a comparison to show that diverges.
Using a similar function, show that diverges.
Exercise & General Exercises