9231. FP2. Integration

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.

Worked Examples

Integrate $\frac{1}{\sqrt{1-x^2}}$ with respect to x
Calculate $\int \frac{4}{4+x^2} dx$
Find the exact value of the integral $\int_0^{1.5} \sqrt{9-x^2} dx$

hyperbolic functions can often be very effective substitutions. For instance, we could calculate $\int \frac{1}{\sqrt{x^2-1}} dx$ 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 $\int \frac{sinh(u)}{sinh(u)}du$ = u + c = cosh^-1x + c

Worked Examples

Using appropriate substitutions (which may be hyperbolic or trigonometric, determine the result of:
- $\int \frac {3}{\sqrt{x^2+4}} dx$
- $\int \frac {3}{\sqrt{9-4x^2}} dx$

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:

Worked Examples

Integrate with respect to x:
- $\frac {1}{x^2+2x+2}$
- $\frac {1}{x^2-4x+7}$
- $\frac {1}{x^2+4x}$

Integrating Hyperbolic Functions

From what we previously saw differentiating hyperbolic functions, it should be clear that $\int sinhx dx = cosh x + c$ and $\int coshx dx = sinh x + c$

To integrate tanh x, we see that $\int tanh dx = \int \frac{sinh x}{cosh x} dx$ and notice that this is of the form $\frac{f'(x)}{f(x)}$ , so integrates to give ln|coshx| + c. Similarly, $\int coth x dx = \int \frac{coshx}{sinhx} dx = ln \mid sinhx \mid + c$ .

Worked Examples

Find a solution for the following integrals:
- $\int 2sinh3x dx$
- $\int \frac{1}{sinhx} dx$
- $\int_{ln2}^{ln3} 3cosh4x dx$

Now let’s integrate inverse hyperbolic functions:

- = $\int cosh^{-1}x \times 1 dx$
- Let u = cosh^-1x and v’ = 1, so $u' = \frac{1}{\sqrt{x^2-1}}$ and v = x, so
- $\int cosh^{-1}x \times 1 dx$ = $xcosh^{-1}x - \int \frac {x}{\sqrt{x^2-1}}$ = $xcosh^{-1}x - \sqrt{x^2-1} + c$

Worked Example

Integrate tanh^-1x with respect to x.

Exercise

Answers

Reduction Formulae

To introduce this concept, let’s consider integrating cos²x using integration by parts:

$I_2 = \int cos^2x dx$
= $\int cosx \times cosx dx$ . Let u = cosx and v’ = cos x, so
$I_2 = cosxsinx - \int (-sinx)(sinx) dx$
$= sinxcosx + \int sin^2x dx$
$= sinxcosx + \int (1-cos^2x) dx$
$= sinxcosx + \int 1 dx - I_2$ , so:
$I_2 = sinxcosx + x + c - I_2$ , which we can rearrange to give:
$2I_2 = sinxcosx + x + c$ , or
$I_2 = \frac{1}{2}(sinxcosx + x) + c$

Worked Example

Given that I₃ = $\int cos^3{x} dx$ , find the result for I₃ using integration by parts.

Now we will try to calculate $I_n = \int cos^nx dx$

$I_n = \int cos^{n-1}x cosx dx$ . Let u=cos^n-1x and v’=cosx, so
$I_n = \cos^{n-1}x sinx - \int (n-1)cos^{n-2}x(-sinx)sinx dx$
= $\cos^{n-1}x sinx +(n-1) \int cos^{n-2}x sin^2x dx$ . Using sin²x=1-cos²x, we get:
= $\cos^{n-1}x sinx +(n-1) \int cos^{n-2}x - (n-1) \int cos^nx$ ,
So $I_n = cos^{n-1}x sinx +(n-1) I_{n-2} - (n-1) I_n$

This rearranges to give the reduction formula valid for n≥2: $I_n = \frac{1}{n}(cos^{n-1}x sinx) + \frac{n-1}{n} I_{n-2}$

Below is an example of how we would use this to calculate $\int cos^4x dx$ for instance, which can be called I₄.

$I_4 = \frac{1}{4}(cos^3xsinx) + \frac{3}{4}I_2$

$I_2 = \frac{1}{2}(cosxsinx) + \frac{1}{2}I_0$

$I_0 = \int cos(0x) dx = \int 1 dx = x$

So $I_4 = \frac{1}{4}(cos^3xsinx) + \frac{3}{4}(\frac{1}{2}(cosxsinx) + \frac{1}{2}x$

$=\frac{1}{4}(cos^3xsinx) + \frac{3}{8}(cosxsinx + x) + c$

Worked Example

If $I_n = \int sin^nx dx$ , show, using integration by parts, that $I_n = \frac{1}{n}(-cosxsin^{n-1}x) + \frac{n-1}{n}I_{n-2}$ for n≥2

We can use the same approach with hyperbolic functions, as shown below for cosh(2x):

$I_n = \int cosh^n2xdx = \int cosh^{n-1}2x cosh 2x dx$ . Let u=cosh^n-12x and v’=cosh2x, so
$I_n = (cosh^{n-1}2x)(\frac{1}{2}sinh2x) - \int ((n-1)cosh^{n-2}2x)(sinh2x)(2)(\frac{1}{2}sinh2x)$
= $\frac{1}{2}cosh^{n-1}2xsinh2x - \int ((n-1)cosh^{n-2}2x)(sinh2x)^{2}$ . We use sinh²2x = cosh²2x – 1 to give:
= $\frac{1}{2}cosh^{n-1}2xsinh2x - \int ((n-1)cosh^{n}2x) +\int((n-1)cosh^{n-2}2x)$ . So
$I_n = \frac{1}{2}cosh^{n-1}2xsinh2x - (n-1)I_n +(n-1)I_{n-2}$ , which rearranges to give:
$I_n = \frac{1}{2n}cosh^{n-1}2xsinh2x + \frac{n-1}{n}I_{n-2}$ , which is valid for all n≥2

Worked Example

If $I_n = \int coth^{n}x dx$ , show that the reduction formula for n≥2 is $I_n = \frac{1}{n-2}(-coth^{n-1}x) + I_n-2$ .

Reduction formulae can also be used with functions that are neither trigonometric nor hyperbolic.

Let’s use this technique to get reduction formulae for:
$I_n = \int x^ne^x dx$ ;
$I_n = \int x^2(1+x^6)^n dx$ . (note that here, once we have used integration by parts, we will need to rewrite x⁸ as x²(1+x⁶-1)).

Worked Example

If $I_n = \int x^n(x^2-1)^9 dx$ , show using integration by parts, that $I_n = \frac{x^{n-1}(x^2-1)^{10}}{n+19} + \frac{n-1}{n+19}I_{n-2}$ for n≥2.

For definite integration, we should use the reduction formula before substituting the limits.

Worked Example

Show that $I_n = \int^{\frac{\pi}{2}}_0 x^ncosx dx$ has reduction formula $I_n = (\frac{\pi}{2})^n - n(n-1)I_{n-2}$ for n≥2 and hence find the exact value of $\int^{\frac{\pi}{2}}_0 x^3cosx dx$

Exercise

Answers

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 $y = \frac{1}{4}x^{\frac{3}{2}}$ 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: $\frac{(\delta s)^2}{(\delta x)^2} = \frac{(\delta x)^2}{(\delta x)^2} + \frac{(\delta y)^2}{(\delta x)^2}$ , so $\frac{(\delta s)^2}{(\delta x)^2} = 1 + \frac{(\delta y)^2}{(\delta x)^2}$ .

Taking the square root of both sides gives: $\frac{\delta s}{\delta x} = \sqrt{1 + \frac{(\delta y)^2}{(\delta x)^2}}$ .

It is by taking the integral at the limit as 𝛿x tends to zero that we thus get the cartesian equation for arc length: $s = \int^{x_2}_{x_1}\sqrt{1+(\frac{dy}{dx})^2} dx$

Worked Example

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: $\frac{(\delta s)^2}{(\delta t)^2} = \frac{(\delta x)^2}{(\delta t)^2} + \frac{(\delta y)^2}{(\delta t)^2}$ .

So following the same steps as before, from this we derive the equation for the length of an arc in parametric form: $s = \int^{t_2}_{t_1}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt$

Worked Example

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=e^𝜽):

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 $\frac{\delta s}{\delta \theta} = \sqrt{ r^2 + (\frac{\delta r}{\delta \theta})^2} d \theta$ , which leads to our formula for the length of an arc in polar form, $s = \int^{\theta_2}_{\theta_1} \sqrt{ r^2 + (\frac{dr}{d \theta})^2} d \theta$

Worked Example

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, $S = \int^{x_2}_{x_1}2 \pi y \sqrt {1 + (\frac{dy}{dx})^2} dx$ ; and
For parametric curves, $S = \int^{t_2}_{t_1}2 \pi y \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dx$ .

Worked Examples

Find the surface area when the curve $y = \frac{1}{3} x^3$ is rotated around the x-axis from x=0 to x=1.

Givent that $y = x^{\frac{1}{2}}$ , 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, $S = \int^{x_2}_{x_1}2 \pi x \sqrt {1 + (\frac{dy}{dx})^2} dx$ ; and
For parametric curves, $S = \int^{t_2}_{t_1}2 \pi x \sqrt {(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dx$ .

Worked Example

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.

Exercise

Answers

Limits of Areas

Let’s think about infinite series, e.g. $\sum^{\infty}_{n=1} f(n)$ .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 $\int^{\infty}_{1} f(x)dx$ . This method can help us identify whether a series converges or diverges.

Consider for instance, the harmonic series, $\sum^{\infty}_{n=1} \frac{1}{n}$ . We can show that this diverges:

$\int^{n}_{1} \frac{1}{n} dx$ = [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 $\sum^{\infty}_{n=1} \frac{1}{n^2}$ . 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 $\frac{1}{2^2} + \frac{1}{3^2} +... + \frac{1}{n^2}$ , so to the area under the curve plus 1 is an upper bound for $\sum^{\infty}_{r=1} \frac{1}{n^2}$ .

So $\sum^{n}_{r=1} \frac{1}{r^2} < 1 + \int^n_1 \frac{1}{x^2}dx$

= $1 + [\frac{-1}{x}]^n_1dx$

= $2 - \frac{1}{n}$ .

In summary, $\sum^{\infty}_{r=1} \frac{1}{r^2} < 2 - \frac{1}{n}$ , and as n->∞ this upper boundary tends towards 2.

We can also use the first graph to identify a lower bound:

$\sum^{n}_{r=1} \frac{1}{r^2} > \int^{n+1}_1 \frac{1}{x^2}dx$ , so

$\sum^{n}_{r=1} \frac{1}{r^2} > 1 - \frac{1}{n+1}$ , so the lower limit as n-> is 1.

Worked Example

(a) Find the upper and lower bounds for $\sum^{n}_{r=1} \frac{1}{r \sqrt{r}}$

(b) Evaluate the infinite sum $\sum^{\infty}_{n=1} \frac{1}{n \sqrt{n}}$ . 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.

Worked Example

Find the upper bound for $\sum^n_{r=2} \frac{1}{rlnr}$ and hence show that $\sum^{\infty}_{n=2} \frac{1}{nlnn}$ diverges.

Comparison Test

Convergence and divergence can also be identified by comparing summations with known intervals. What would be a suitable comparison for the previous example?

Worked Examples

Use a comparison to show that $\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}$ diverges.