771 lines
21 KiB
Plaintext
771 lines
21 KiB
Plaintext
# Substitution
|
||
|
||
This section uses these add-on packages:
|
||
|
||
```julia
|
||
using CalculusWithJulia
|
||
using Plots
|
||
using SymPy
|
||
|
||
```
|
||
|
||
```julia; echo=false; results="hidden"
|
||
using CalculusWithJulia.WeaveSupport
|
||
|
||
const frontmatter = (
|
||
title = "Substitution",
|
||
description = "Calculus with Julia: Substitution",
|
||
tags = ["CalculusWithJulia", "integrals", "substitution"],
|
||
);
|
||
nothing
|
||
```
|
||
|
||
----
|
||
|
||
The technique of $u$-[substitution](https://en.wikipedia.org/wiki/Integration_by_substitution) is derived from reversing the chain rule: $[f(g(x))]' = f'(g(x)) g'(x)$.
|
||
|
||
Suppose that $g$ is continuous and $u(x)$ is differentiable with ``u'(x)`` being Riemann integrable. Then both these integrals are defined:
|
||
|
||
```math
|
||
\int_a^b g(u(t)) \cdot u'(t) dt, \quad \text{and}\quad \int_{u(a)}^{u(b)} g(x) dx.
|
||
```
|
||
|
||
We wish to show they are equal.
|
||
|
||
Let $G$ be an antiderivative of $g$, which exists as $g$ is assumed to be continuous. (By the Fundamental Theorem part I.) Consider the composition $G \circ u$. The chain rule gives:
|
||
|
||
```math
|
||
[G \circ u]'(t) = G'(u(t)) \cdot u'(t) = g(u(t)) \cdot u'(t).
|
||
```
|
||
|
||
So,
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int_a^b g(u(t)) \cdot u'(t) dt &= \int_a^b (G \circ u)'(t) dt\\
|
||
&= (G\circ u)(b) - (G\circ u)(a) \quad\text{(the FTC, part II)}\\
|
||
&= G(u(b)) - G(u(a)) \\
|
||
&= \int_{u(a)}^{u(b)} g(x) dx. \quad\text{(the FTC part II)}
|
||
\end{align*}
|
||
```
|
||
|
||
|
||
That is, this substitution formula applies:
|
||
|
||
> ``\int_a^b g(u(x)) u'(x) dx = \int_{u(a)}^{u(b)} g(x) dx.``
|
||
|
||
Further, for indefinite integrals,
|
||
|
||
> ``\int f(g(x)) g'(x) dx = \int f(u) du.``
|
||
|
||
|
||
|
||
We have seen a special case of substitution where $u(x) = x-c$ in the formula $\int_{a-c}^{b-c} g(x) dx= \int_a^b g(x-c)dx$.
|
||
|
||
|
||
|
||
The main use of this is to take complicated things inside of the function $g$ out of the function (the $u(x)$) by renaming them, then accounting for the change of name.
|
||
|
||
Some examples are in order.
|
||
|
||
Consider:
|
||
|
||
```math
|
||
\int_0^{\pi/2} \cos(x) e^{\sin(x)} dx.
|
||
```
|
||
|
||
Clearly the $\sin(x)$ inside the exponential is an issue. If we let $u(x) = \sin(x)$, then $u'(x) = \cos(x)$, and this becomes
|
||
|
||
```math
|
||
\int_0^2 u\prime(x) e^{u(x)} dx =
|
||
\int_{u(0)}^{u(\pi/2)} e^x dx = e^x \big|_{\sin(0)}^{\sin(\pi/2)} = e^1 - e^0.
|
||
```
|
||
|
||
This all worked, as the problem was such that it was more or less obvious what to choose for $u$ and $G$.
|
||
|
||
### Integration by substitution
|
||
|
||
The process of identifying the result of the chain rule in the
|
||
function to integrate is not automatic, but rather a bit of an art. The basic
|
||
step is to try some values and hope one works. Typically, this is taught by
|
||
"substituting" in some value for part of the expression (basically the
|
||
$u(x)$) and seeing what happens.
|
||
|
||
In the above problem, $\int_0^{\pi/2} \cos(x) e^{\sin(x)} dx$, we
|
||
might just rename $\sin(x)$ to be $u$ (suppressing the "of $x$
|
||
part). Then we need to rewrite the "$dx$" part of the integral. We
|
||
know in this case that $du/dx = \cos(x)$. In terms of differentials,
|
||
this gives $du = \cos(x) dx$. But this allows us to substitute in with
|
||
$u$ and $du$ as is possible:
|
||
|
||
```math
|
||
\int_0^{\pi/2} \cos(x) e^{\sin(x)} dx = \int_0^{\pi/2} e^{\sin(x)} \cdot \cos(x) dx = \int_{u(0)}^{u(\pi)} e^u du.
|
||
```
|
||
|
||
----
|
||
|
||
Let's illustrate with a new problem: $\int_0^2 4x e^{x^2} dx$.
|
||
|
||
Again, we see that the $x^2$ inside the exponential is a complication. Letting $u = x^2$ we have $du = 2x dx$. We have $4xdx$ in the
|
||
original problem, so we will end up with $2du$:
|
||
|
||
```math
|
||
\int_0^2 4x e^{x^2} dx = 2\int_0^2 e^{x^2} \cdot 2x dx = 2\int_{u(0)}^{u(2)} e^u du = 2 \int_0^4 e^u du =
|
||
2 e^u\big|_{u=0}^4 = 2(e^4 - 1).
|
||
```
|
||
|
||
----
|
||
|
||
Consider now $\int_0^1 2x^2 \sqrt{1 + x^3} dx$. Here we see that the $1 + x^3$ makes the square root term complicated. If we call this $u$, then what is $du$? Clearly, $du = 3x^2 dx$, or $(1/3)du = x^2 dx$, so we can rewrite this as:
|
||
|
||
```math
|
||
\int_0^1 2x^2 \sqrt{1 + x^3} dx = \int_{u(0)}^{u(1)} 2 \sqrt{u} (1/3) du = 2/3 \cdot \frac{u^{3/2}}{3/2} \big|_1^2 =
|
||
\frac{4}{9} \cdot(2^{3/2} - 1).
|
||
```
|
||
|
||
|
||
----
|
||
|
||
Consider $\int_0^{\pi} \cos(x)^3 \sin(x) dx$. The $\cos(x)$ function inside the $x^3$ function is complicated. We let $u(x) = \cos(x)$ and see what that implies: $du = \sin(x) dx$, which we see is part of the question. So the above becomes:
|
||
|
||
```math
|
||
\int_0^{\pi} \cos(x)^3 \sin(x) dx = \int_{u(0)}^{u(\pi)} u^3 du= \frac{u^4}{4}\big|_0^0 = 0.
|
||
```
|
||
|
||
Changing limits leaves the two endpoints the same, which means the
|
||
total area after substitution is $0$. A graph of this function shows
|
||
that about $\pi/2$ the function has odd-like symmetry, so the answer of $0$ is supported by the plot:
|
||
|
||
|
||
```julia;hold=true
|
||
f(x) = cos(x)^3 * sin(x)
|
||
plot(f, 0, 1pi)
|
||
```
|
||
|
||
|
||
|
||
----
|
||
|
||
Consider $\int_1^e \log(x)/x dx$. There isn't really an "inside" function here, but instead just a tricky $\log(x)$. If we let $u=\log(x)$, what happens? We get $du = 1/x \cdot dx$, which we see present in the original. So with this, we have:
|
||
|
||
```math
|
||
\int_1^e \frac{\log(x)}{x} dx = \int_{u(1)}^{u(e)} u du = \frac{u^2}{2}\big|_0^1 = \frac{1}{2}.
|
||
```
|
||
|
||
##### Example: Transformations
|
||
|
||
We say that the area intrinsically discussed in the definite integral
|
||
$A=\int_a^b f(x-c) dx$ is unaffected by shifts, in that $A =
|
||
\int_{a-c}^{b-c} f(x) dx$. What about more general transformations?
|
||
For example: if $g(x) = (1/h) \cdot f((x-c)/h)$ for values $c$ and $h$ what is
|
||
the integral over $a$ to $b$ in terms of the function $f(x)$?
|
||
|
||
If $A = \int_a^b (1/h) \cdot f((x-c)/h) dx$ then we let $u = (x-c)/h$. With this, $du = 1/h \cdot dx$. This allows a straight substitution:
|
||
|
||
```math
|
||
A = \int_a^b \frac{1}{h} f(\frac{x-c}{h}) dx = \int_{(a-c)/h}^{(b-c)/h} f(u) du.
|
||
```
|
||
|
||
So the answer is: the area under the transformed function over $a$ to $b$
|
||
is the area of the function over the transformed region.
|
||
|
||
|
||
For example, consider the "hat" function $f(x) = 1 - \lvert x \rvert $
|
||
when $-1 \leq x \leq 1$ and $0$ otherwise. The area under $f$ is just
|
||
$1$ - the graph forms a triangle with base of length $2$ and height
|
||
$1$. If we take any values of $c$ and $h$, what do we find for the
|
||
area under the curve of the transformed function?
|
||
|
||
Let $u(x) = (x-c)/h$ and $g(x) = h f(u(x))$. Then, as $du = 1/h dx$
|
||
|
||
```math
|
||
\begin{align}
|
||
\int_{c-h}^{c+h} g(x) dx
|
||
&= \int_{c-h}^{c+h} h f(u(x)) dx\\
|
||
&= \int_{u(c-h)}^{u(c+h)} f(u) du\\
|
||
&= \int_{-1}^1 f(u) du\\
|
||
&= 1.
|
||
\end{align}
|
||
```
|
||
|
||
So the area of this transformed function is still $1$. The shifting by
|
||
$c$ we know doesn't effect the area, the scaling by $h$ inside of $f$
|
||
does, but is balanced out by the multiplication by $h$ outside of $f$.
|
||
|
||
##### Example: Speed versus velocity
|
||
|
||
The "velocity" of an object includes a sense of direction in addition to the sense of magnitude. The "speed" just includes the sense of magnitude. Speed is always non-negative, whereas velocity is a signed quantity.
|
||
|
||
As mentioned previously, position is the integral of velocity, as expressed precisely through this equation:
|
||
|
||
```math
|
||
x(t) = \int_0^t v(u) du - x(0).
|
||
```
|
||
|
||
What is the integral of speed?
|
||
|
||
If $v(t)$ is the velocity, the $s(t) = \lvert v(t) \rvert$ is the speed. If integrating either $s(t)$ or $v(t)$, the integrals would agree when $v(t) \geq 0$. However, when $v(t) \leq 0$, the position back tracks so $x(t)$ decreases, where the integral of $s(t)$ would only increase.
|
||
|
||
This integral
|
||
|
||
```math
|
||
td(t) = \int_0^t s(u) du = \int_0^t \lvert v(u) \rvert du,
|
||
```
|
||
|
||
Gives the *total distance* traveled.
|
||
|
||
To illustrate with a simple example, if a car drives East for one hour
|
||
at 60 miles per hour, then heads back West for an hour at 60 miles per
|
||
hour, the car's position after one hour is $x(2) = x(0)$, with a change in position $x(2) - x(0) = 0$. Whereas, the
|
||
total distance traveled is $120$ miles. (Gas is paid on total
|
||
distance, not change in position!). What are the formulas for speed
|
||
and velocity? Clearly $s(t) = 60$, a constant, whereas here $v(t) =
|
||
60$ for $0 \leq t \leq 1$ and $-60$ for $1 < t \leq 2$.
|
||
|
||
|
||
|
||
Suppose $v(t)$ is given by $v(t) = (t-2)^3/3 - 4(t-2)/3$. If $x(0)=0$
|
||
Find the position after 3 time units and the total distance traveled.
|
||
|
||
|
||
We let $u(t) = t - 2$ so $du=dt$. The position is given by
|
||
|
||
```math
|
||
\int_0^3 ((t-2)^3/3 - 4(t-2)/3) dt = \int_{u(0)}^{u(3)} (u^3/3 - 4/3 u) du =
|
||
(\frac{u^4}{12} - \frac{4}{3}\frac{u^2}{2}) \big|_{-2}^1 = \frac{3}{4}.
|
||
```
|
||
|
||
The speed is similar, but we have to work harder:
|
||
|
||
|
||
```math
|
||
\int_0^3 \lvert v(t) \rvert dt = \int_0^3 \lvert ((t-2)^3/3 - 4(t-2)/3) \rvert dt =
|
||
\int_{-2}^1 \lvert u^3/3 - 4u/3 \rvert du.
|
||
```
|
||
|
||
But $u^3/3 - 4u/3 = (1/3) \cdot u(u-1)(u+2)$, so between $-2$ and $0$
|
||
it is positive and between $0$ and $1$ negative, so this integral is:
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int_{-2}^0 (u^3/3 - 4u/3 ) du + \int_{0}^1 -(u^3/3 - 4u/3) du
|
||
&= (\frac{u^4}{12} - \frac{4}{3}\frac{u^2}{2}) \big|_{-2}^0 - (\frac{u^4}{12} - \frac{4}{3}\frac{u^2}{2}) \big|_{0}^1\\
|
||
&= \frac{4}{3} - -\frac{7}{12}\\
|
||
&= \frac{23}{12}.
|
||
\end{align*}
|
||
```
|
||
|
||
##### Example
|
||
|
||
In probability, the normal distribution plays an outsized role. This distribution is characterized by a family of *density* functions:
|
||
|
||
```math
|
||
f(x; \mu, \sigma) = \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma} \exp(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2).
|
||
```
|
||
|
||
Integrals involving this function are typically transformed by substitution. For example:
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int_a^b f(x; \mu, \sigma) dx
|
||
&= \int_a^b \frac{1}{\sqrt{2\pi}}\frac{1}{\sigma} \exp(-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2) dx \\
|
||
&= \int_{u(a)}^{u(b)} \frac{1}{\sqrt{2\pi}} \exp(-\frac{1}{2}u^2) du \\
|
||
&= \int_{u(a)}^{u(b)} f(u; 0, 1) du,
|
||
\end{align*}
|
||
```
|
||
|
||
where ``u = (x-\mu)/\sigma``, so ``du = (1/\sigma) dx``.
|
||
|
||
This shows that integrals involving a normal density with parameters ``\mu`` and ``\sigma`` can be computed using the *standard* normal density with ``\mu=0`` and ``\sigma=1``. Unfortunately, there is no elementary antiderivative for ``\exp(-u^2/2)``, so integrals for the standard normal must be numerically approximated.
|
||
|
||
There is a function `erf` in the `SpecialFunctions` package (which is loaded by `CalculusWithJulia`) that computes:
|
||
|
||
```math
|
||
\int_0^x \frac{2}{\sqrt{\pi}} \exp(-t^2) dt
|
||
```
|
||
|
||
A further change of variables by ``t = u/\sqrt{2}`` (with ``\sqrt{2}dt = du``) gives:
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int_a^b f(x; \mu, \sigma) dx &=
|
||
\int_{t(u(a))}^{t(u(b))} \frac{\sqrt{2}}{\sqrt{2\pi}} \exp(-t^2) dt\\
|
||
&= \frac{1}{2} \int_{t(u(a))}^{t(u(b))} \frac{2}{\sqrt{\pi}} \exp(-t^2) dt
|
||
\end{align*}
|
||
```
|
||
|
||
Up to a factor of ``1/2`` this is `erf`.
|
||
|
||
So we would have, for example, with ``\mu=1``,``\sigma=2`` and ``a=1`` and ``b=3`` that:
|
||
|
||
```math
|
||
\begin{align*}
|
||
t(u(a)) &= (1 - 1)/2/\sqrt{2} = 0\\
|
||
t(u(b)) &= (3 - 1)/2/\sqrt{2} = \frac{1}{\sqrt{2}}\\
|
||
\int_1^3 f(x; 1, 2)
|
||
&= \frac{1}{2} \int_0^{1/\sqrt{2}} \frac{2}{\sqrt{\pi}} \exp(-t^2) dt.
|
||
\end{align*}
|
||
```
|
||
|
||
Or
|
||
|
||
```julia
|
||
1/2 * erf(1/sqrt(2))
|
||
```
|
||
|
||
!!! note "The `Distributions` package"
|
||
|
||
The above calculation is for illustration purposes. The add-on package `Distributions` makes much quicker work of such a task for the normal distribution and many other distributions from probability and statistics.
|
||
|
||
## SymPy and substitution
|
||
|
||
The `integrate` function in `SymPy` can handle most problems which involve substitution. Here are a few examples:
|
||
|
||
|
||
* This integral, $\int_0^2 4x/\sqrt{x^2 +1}dx$, involves a substitution for $x^2 + 1$:
|
||
|
||
```julia;
|
||
@syms x::real t::real
|
||
integrate(4x / sqrt(x^2 + 1), (x, 0, 2))
|
||
```
|
||
|
||
|
||
* This integral, $\int_e^{e^2} 1/(x\log(x)) dx$ involves a substitution of $u=\log(x)$. Here we see the answer:
|
||
|
||
```julia; hold=true
|
||
f(x) = 1/(x*log(x))
|
||
integrate(f(x), (x, sympy.E, sympy.E^2))
|
||
```
|
||
|
||
(We used `sympy.E)` - and not `e` - to avoid any conversion to floating point, which could yield an inexact answer.)
|
||
|
||
|
||
The antiderivative is interesting here; it being an *iterated* logarithm.
|
||
|
||
```julia;
|
||
integrate(1/(x*log(x)), x)
|
||
```
|
||
|
||
|
||
|
||
### Failures...
|
||
|
||
|
||
Not every integral problem lends itself to solution by
|
||
substitution. For example, we can use substitution to evaluate the
|
||
integral of $xe^{-x^2}$, but for $e^{-x^2}$ or $x^2e^{-x^2}$. The
|
||
first has no familiar antiderivative, the second is done by a
|
||
different technique.
|
||
|
||
Even when substitution can be used, `SymPy` may not be able to
|
||
algorithmically identify it. The main algorithm used can determine if
|
||
expressions involving rational functions, radicals, logarithms, and
|
||
exponential functions is integrable. Missing from this list are
|
||
absolute values.
|
||
|
||
|
||
For some such problems, we can help `SymPy` out - by breaking the integral into pieces where we know the sign of the expression.
|
||
|
||
For substitution problems, we can also help out. For example, to find an antiderivative for
|
||
|
||
```math
|
||
\int(1 + \log(x)) \sqrt{1 + (x\log(x))^2} dx
|
||
```
|
||
|
||
A quick attempt with `SymPy` turns up nothing:
|
||
|
||
|
||
```julia;
|
||
𝒇(x) = (1 + log(x)) * sqrt(1 + (x*log(x))^2 )
|
||
integrate(𝒇(x), x)
|
||
```
|
||
|
||
But were we to try $u=x\log(x)$, we'd see that this simplifies to $\int \sqrt{1 + u^2} du$, which has some hope of having an antiderivative.
|
||
|
||
We can help `SymPy` out by substitution:
|
||
|
||
```julia;
|
||
u(x) = x * log(x)
|
||
@syms w dw
|
||
ex = 𝒇(x)
|
||
ex₁ = ex(u(x) => w, diff(u(x),x) => dw)
|
||
```
|
||
|
||
This verifies the above. Can it be integrated in `w`? The "`dw`" is only for familiarity, `SymPy` doesn't use this, so we set it to 1 then integrate:
|
||
|
||
```julia;
|
||
ex₂ = ex₁(dw => 1)
|
||
ex₃ = integrate(ex₂, w)
|
||
```
|
||
|
||
Finally, we put back in the `u(x)` to get an antiderivative.
|
||
|
||
```julia;
|
||
ex₃(w => u(x))
|
||
```
|
||
|
||
!!! note
|
||
Lest it be thought this is an issue with `SymPy`, but not other
|
||
systems, this example was [borrowed](http://faculty.uml.edu/jpropp/142/Integration.pdf) from an
|
||
illustration for helping Mathematica.
|
||
|
||
|
||
## Trigonometric substitution
|
||
|
||
Wait, in the last example an antiderivative for $\sqrt{1 + u^2}$ was found. But how? We haven't discussed this yet.
|
||
|
||
|
||
This can be found using *trigonometric* substitution. In this example,
|
||
we know that $1 + \tan(\theta)^2$ simplifies to $\sec(\theta)^2$, so
|
||
we might *try* a substitution of $\tan(u)=x$. This would simplify
|
||
$\sqrt{1 + x^2}$ to $\sqrt{1 + \tan(u)^2} = \sqrt{\sec(u)^2}$ which is
|
||
$\lvert \sec(u) \rvert$. What of $du$? The chain rule gives
|
||
$\sec(u)^2du = dx$. In short we get:
|
||
|
||
```math
|
||
\int \sqrt{1 + x^2} dx = \int \sec(u)^2 \lvert \sec(u) \rvert du = \int \sec(u)^3 du,
|
||
```
|
||
|
||
if we know ``\sec(u) \geq 0``.
|
||
|
||
|
||
This leaves still the question of integrating $\sec(u)^3$, which we aren't (yet) prepared to discuss, but we see that this type of substitution can re-express an integral in a new way that may pay off.
|
||
|
||
#### Examples
|
||
|
||
Let's see some examples where a trigonometric substitution is all that is needed.
|
||
|
||
##### Example
|
||
|
||
Consider $\int 1/(1+x^2) dx$. This is an antiderivative of some function, but if that isn't observed, we might notice the $1+x^2$ and try to simplify that. First, an attempt at a $u$-substitution:
|
||
|
||
|
||
Letting $u = 1+x^2$ we get $du = 2xdx$ which gives $\int (1/u) (2x) du$. We aren't able to address the "$2x$" part successfully, so this attempt is for naught.
|
||
|
||
|
||
Now we try a trigonometric substitution, taking advantage of the identity $1+\tan(x)^2 = \sec(x)^2$. Letting $\tan(u) = x$ yields $\sec(u)^2 du = dx$ and we get:
|
||
|
||
```math
|
||
\int \frac{1}{1+x^2} dx = \int \frac{1}{1 + \tan(u)^2} \sec(u)^2 du = \int 1 du = u.
|
||
```
|
||
|
||
But $\tan(u) = x$, so in terms of $x$, an antiderivative is just $\tan^{-1}(x)$, or the arctangent. Here we verify with `SymPy`:
|
||
|
||
```julia;
|
||
integrate(1/(1+x^2), x)
|
||
```
|
||
|
||
The general form allows ``a^2 + (bx)^2`` in the denominator (squared so both are positive and the answer is nicer):
|
||
|
||
```julia; hold=true
|
||
@syms a::real, b::real, x::real
|
||
integrate(1 / (a^2 + (b*x)^2), x)
|
||
```
|
||
|
||
|
||
##### Example
|
||
|
||
The expression $1-x^2$ can be attacked by the substitution $\sin(u) =x$ as then $1-x^2 = 1-\cos(u)^2 = \sin(u)^2$. Here we see this substitution being used successfully:
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int \frac{1}{\sqrt{9 - x^2}} dx &= \int \frac{1}{\sqrt{9 - (3\sin(u))^2}} \cdot 3\cos(u) du\\
|
||
&=\int \frac{1}{3\sqrt{1 - \sin(u)^2}}\cdot3\cos(u) du \\
|
||
&= \int du \\
|
||
&= u \\
|
||
&= \sin^{-1}(x/3).
|
||
\end{align*}
|
||
```
|
||
|
||
|
||
Further substitution allows the following integral to be solved for an antiderivative:
|
||
|
||
```julia; hold=true
|
||
@syms a::real, b::real
|
||
integrate(1 / sqrt(a^2 - b^2*x^2), x)
|
||
```
|
||
|
||
##### Example
|
||
|
||
The expression $x^2 - 1$ is a bit different, this lends itself to $\sec(u) = x$ for a substitution, for $\sec(u)^2 - 1 = \tan(u)^2$. For example, we try $\sec(u) = x$ to integrate:
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int \frac{1}{\sqrt{x^2 - 1}} dx &= \int \frac{1}{\sqrt{\sec(u)^2 - 1}} \cdot \sec(u)\tan(u) du\\
|
||
&=\int \frac{1}{\tan(u)}\sec(u)\tan(u) du\\
|
||
&= \int \sec(u) du.
|
||
\end{align*}
|
||
```
|
||
|
||
|
||
This doesn't seem that helpful, but the antiderivative to $\sec(u)$ is
|
||
$\log\lvert (\sec(u) + \tan(u))\rvert$, so we can proceed to get:
|
||
|
||
```math
|
||
\begin{align*}
|
||
\int \frac{1}{\sqrt{x^2 - 1}} dx &= \int \sec(u) du\\
|
||
&= \log\lvert (\sec(u) + \tan(u))\rvert\\
|
||
&= \log\lvert x + \sqrt{x^2-1} \rvert.
|
||
\end{align*}
|
||
```
|
||
|
||
SymPy gives a different representation using the arccosine:
|
||
|
||
```julia; hold=true
|
||
@syms a::positive, b::positive, x::real
|
||
integrate(1 / sqrt(a^2*x^2 - b^2), x)
|
||
```
|
||
|
||
|
||
|
||
##### Example
|
||
|
||
The equation of an ellipse is $x^2/a^2 + y^2/b^2 = 1$. Suppose $a,b>0$. The area under
|
||
the function $b \sqrt{1 - x^2/a^2}$ between $-a$ and $a$
|
||
will then be half the area of the ellipse. Find the area enclosed by
|
||
the ellipse.
|
||
|
||
We need to compute:
|
||
|
||
```math
|
||
2\int_{-a}^a b \sqrt{1 - x^2/a^2} dx =
|
||
4 b \int_0^a\sqrt{1 - x^2/a^2} dx.
|
||
```
|
||
|
||
Letting $\sin(u) = x/a$ gives $a\cos(u)du = dx$ and an antiderivative is found with:
|
||
|
||
```math
|
||
4 b \int_0^a \sqrt{1 - x^2/a^2} dx = 4b \int_0^{\pi/2} \sqrt{1-u^2} a \cos(u) du
|
||
= 4ab \int_0^{\pi/2} \cos(u)^2 du
|
||
```
|
||
|
||
The identify $\cos(u)^2 = (1 + \cos(2u))/2$ makes this tractable:
|
||
|
||
```math
|
||
\begin{align*}
|
||
4ab \int \cos(u)^2 du
|
||
&= 4ab\int_0^{\pi/2}(\frac{1}{2} + \frac{\cos(2u)}{2}) du\\
|
||
&= 4ab(\frac{1}{2}u + \frac{\sin(2u)}{4})\big|_0^{\pi/2}\\
|
||
&= 4ab (\pi/4 + 0) = \pi ab.
|
||
\end{align*}
|
||
```
|
||
|
||
Keeping in mind that that a circle with radius $a$ is an ellipse with
|
||
$b=a$, we see that this gives the correct answer for a circle.
|
||
|
||
## Questions
|
||
|
||
###### Question
|
||
|
||
For $\int \sin(x) \cos(x) dx$, let $u=\sin(x)$. What is the resulting substitution?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``\\int u du``",
|
||
"``\\int u (1 - u^2) du``",
|
||
"``\\int u \\cos(x) du``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
For $\int \tan(x)^4 \sec(x)2 dx$ what $u$-substitution makes this easy?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``u=\\tan(x)``",
|
||
"``u=\\tan(x)^4``",
|
||
"``u=\\sec(x)``",
|
||
"``u=\\sec(x)^2``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
For $\int x \sqrt{x^2 - 1} dx$ what $u$ substitution makes this easy?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``u=x^2 - 1``",
|
||
"``u=x^2``",
|
||
"``u=\\sqrt{x^2 - 1}``",
|
||
"``u=x``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
###### Question
|
||
|
||
For $\int x^2(1-x)^2 dx$ will the substitution $u=1-x$ prove effective?
|
||
|
||
```julia; hold=true; echo=false
|
||
yesnoq("no")
|
||
```
|
||
|
||
What about expanding the factored polynomial to get a fourth degree polynomial, will this prove effective?
|
||
|
||
```julia; hold=true; echo=false
|
||
yesnoq("yes")
|
||
```
|
||
|
||
###### Question
|
||
|
||
For $\int (\log(x))^3/x dx$ the substitution $u=\log(x)$ reduces this to what?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``\\int u^3 du``",
|
||
"``\\int u du``",
|
||
"``\\int u^3/x du``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
###### Question
|
||
|
||
For $\int \tan(x) dx$ what substitution will prove effective?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``u=\\cos(x)``",
|
||
"``u=\\sin(x)``",
|
||
"``u=\\tan(x)``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
Integrating $\int_0^1 x \sqrt{1 - x^2} dx$ can be done by using the $u$-substitution $u=1-x^2$. This yields an integral
|
||
|
||
```math
|
||
\int_a^b \frac{-\sqrt{u}}{2} du.
|
||
```
|
||
|
||
What are $a$ and $b$?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``a=0,~ b=1``",
|
||
"``a=1,~ b=0``",
|
||
"``a=0,~ b=0``",
|
||
"``a=1,~ b=1``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
The integral $\int \sqrt{1 - x^2} dx$ lends itself to what substitution?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``\\sin(u) = x``",
|
||
"``\\tan(u) = x``",
|
||
"``\\sec(u) = x``",
|
||
"``u = 1 - x^2``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
The integral $\int x/(1+x^2) dx$ lends itself to what substitution?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``u = 1 + x^2``",
|
||
"``\\sin(u) = x``",
|
||
"``\\tan(u) = x``",
|
||
"``\\sec(u) = x``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
The integral $\int dx / \sqrt{1 - x^2}$ lends itself to what substitution?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``\\sin(u) = x``",
|
||
"``\\tan(u) = x``",
|
||
"``\\sec(u) = x``",
|
||
"``u = 1 - x^2``"
|
||
]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
|
||
|
||
###### Question
|
||
|
||
The integral $\int dx / \sqrt{x^2 - 16}$ lends itself to what substitution?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``4\\sec(u) = x``",
|
||
"``\\sec(u) = x``",
|
||
"``4\\sin(u) = x``",
|
||
"``\\sin(u) = x``"]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
The integral $\int dx / (a^2 + x^2)$ lends itself to what substitution?
|
||
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
"``\\tan(u) = x``",
|
||
"``\\tan(u) = x``",
|
||
"``a\\sec(u) = x``",
|
||
"``\\sec(u) = x``"]
|
||
answ = 1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
The integral $\int_{1/2}^1 \sqrt{1 - x^2}dx$ can be approached with the substitution $\sin(u) = x$ giving:
|
||
|
||
```math
|
||
\int_a^b \cos(u)^2 du.
|
||
```
|
||
|
||
What are $a$ and $b$?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices =[
|
||
"``a=\\pi/6,~ b=\\pi/2``",
|
||
"``a=\\pi/4,~ b=\\pi/2``",
|
||
"``a=\\pi/3,~ b=\\pi/2``",
|
||
"``a=1/2,~ b= 1``"
|
||
]
|
||
answ =1
|
||
radioq(choices, answ)
|
||
```
|
||
|
||
###### Question
|
||
|
||
How would we verify that $\log\lvert (\sec(u) + \tan(u))\rvert$ is an antiderivative for $\sec(u)$?
|
||
|
||
```julia; hold=true; echo=false
|
||
choices = [
|
||
L"We could differentiate $\sec(u)$.",
|
||
L"We could differentiate $\log\lvert (\sec(u) + \tan(u))\rvert$ "]
|
||
answ = 2
|
||
radioq(choices, answ)
|
||
```
|