# import all python add-ons etc that will be needed later on
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from sympy import *
init_printing() # allows printing of SymPy results in typeset maths format
plt.rcParams.update({'font.size': 16}) # set font size for plots
(a) $I=\int\cos^3(x)dx$. Trying $u=\sin(x)$, then $du=\cos(x) dx$ producing
$$\displaystyle I=\int\cos^2(x)du=\int(1-u^2)du=u-u^3/3=\sin(x)-\sin^3(x)/2+c$$where $c$ is a constant and $\cos^2+\sin^2=1$ was used.
(b) $\displaystyle I=\int\frac{3x}{(5+3x)^4}dx$. Guessing $u=5+3x$ then $du=dx/3$ and substituting gives
$$\displaystyle I=\int\frac{u-5}{3u^4}du =-\frac{1}{6u^2}+\frac{5}{9u^3}=-\frac{9x+5}{18(5+3x)}+c$$and eqn 6 was used.
(c) $\displaystyle I=\int\frac{e^\sqrt{x}}{\sqrt{x}}dx$. It seems obvious to try $u=\sqrt{x}$ and $du=dx/(2\sqrt{x})$. Substituting gives
$$\displaystyle I=\int \frac{e^u}{2} du= 2e^\sqrt{x}+c$$(d) $I=\int\cot(ax)dx$. Recognising that $\cot=1/\tan=\cos/\sin$ gives $\displaystyle I=\int\frac{\cos(x)}{\sin(x)}dx$ and the numerator (on top) is $1/a$ times the derivative of $\sin(ax)$. Therefore, using equation 12,
$$\displaystyle I=\int\cot(ax)dx = \frac{1}{a}\ln\left(\sin(ax)\right)+c$$Using Sympy as a check
x, a = symbols('x, a',positive =True)
integrate(cot(a*x),x)
(e) $\displaystyle I=\int \frac{x^2}{8+x^3}dx$. In this case the numerator is 1/3 of the derivative of the denominator, therefore, using equation 12,
$$\displaystyle I=\int \frac{x^2}{8+x^3}dx=\frac{1}{3}\ln(8+x^3)$$(f) $I=\int e^{ax}(1-e^{ax})^{-1}dx$. This also has the form of equation 12 with the derivative as numerator therefore, $I=-\ln(1-e^{ax})/a$.
Using the substitution $x=a\sin(u)$ and $dx=a\cos(u)du$ gives
$$\displaystyle I=\int\frac{1}{\sqrt{a^2-x^2}}dx=\int\frac{\cos(u)}{\sin(u)}du = u=\sin^{-1}\left( \frac{x}{a} \right)$$If a computer algebra application was used the answer $\displaystyle \arctan\left( \frac{a}{\sqrt{a^2-x^2}} \right)$ may be produced: $\arctan() \equiv \tan^{-1}()$. The inverse sine and tangents are related easily with a right-angled triangle. For example, let $y = \sin^{-1}(x/a)$ which is the same as writing $\sin(y) = x/a$. As sine is opposite / hypotenuse in a right-angled triangle, the adjacent side has length $\sqrt{a^2 - x^2}$ and therefore $\tan(y)=x/\sqrt{a^2-x^2}$ producing $\displaystyle \sin^{-1}(x/a)=\tan^{-1}(x/\sqrt{a^2-x^2})$
Figure 41. Trig construction. ____
(a) Let $u = x^2$ and thus $du = 2xdx$ the integral can pleasingly be simplified to
$$\displaystyle I=\frac{1}{2}\int e^{-u}du =-\frac{1}{2}e^{-u}=-\frac{1}{2}e^{-x^2} +c$$(b) Performing the definite integral
$$\displaystyle I=\int_0^x xe^{-x^2}dx=-\frac{e^{-x^2}}{2}\bigg|_0^x= \frac{1-e^{-x^2}}{2} \tag{96}$$Note that this result is only valid is $x\ge 0$, why is this?
Figure 42. Plot of the function $xe^{-x^2}$ and its integral $(1-e^{-x^2})/2 $ vs $x$ (red line) showing an error for the integral when $x \lt 0$, dashed red line. The height of the (red) curve when $x \ge 0$ is the value of the integral. ____
Clearly the integration is incorrect when $x \lt 0$ because the function $xe^{-x^2}$ here is always negative so the integral, as the area under the curve, must also be negative. What has gone wrong? Well, it is because $x$ has been allowed to get smaller than the lower limit in the integral when plotting equation 96. When this happens equation 4 must be used, or the order of integration should be reversed,
$$\displaystyle I=\int_{-x}^0 xe^{-x^2}dx=-\frac{e^{-x^2}}{2}\bigg|_{-x}^0= -\frac{1-e^{-x^2}}{2} $$and the only difference is a change of sign. Now this makes everything correct as shown in Figure 43 .
Expanding the fraction gives two simpler integrals
$$\displaystyle I=\int\frac{x-2}{x\sqrt{x+3}}dx=\int\frac{1}{\sqrt{x+3}}dx=-\int\frac{2}{x\sqrt{x+3}}dx$$In expressions of this form always try substituting for the square root, $u=\sqrt{x+3}$, then $dy=dx/(2\sqrt{x+3})$ and the first integral is $I_1=\int2du=2u=2\sqrt{x+3}$. The second is $I_2=\displaystyle -\int\frac{2}{u^2-3}du$, which is a standard integral, see section 2.13. The result is
$$\displaystyle I_2=\frac{4}{\sqrt{3}}\tanh^{-1}\left(\frac{\sqrt{x+3}}{\sqrt{3}} \right)+c $$Some computer algebra calculations may give the $\tanh^{-1}()$ fraction upside down compared to this answer. In this case the two are related as $\tanh^{-1}(w)=\tanh^{-1}(-1/w)+\pi/2$.
Guessing at the substitution $u = x^3$ to get a term in $x^2dx$ as $du = 3x^2dx$. Changing the limits when $x = -1,\, u = -1$, and when $x = 2, \,u = 8$. The substituted integral can be solved by looking at the table of integrals (Section 2.13),
$$\displaystyle I=\frac{1}{3}\int_{-1}^8 \frac{1}{16+u^2}du=\frac{1}{12}\tan^{-1}\left(\frac{u}{4}\right)\bigg|_{-1}^8=\frac{1}{12}\left(\tan^{-1}(2)+\tan^{-1}\left(\frac{1}{4}\right) \right)$$This integration does not look tractable, but try $u = \sin^{-1}(ax)$, and $dt = dx$ so that $t = x$. Evaluating gives
$$\displaystyle I=\int\sin^{-1}(ax) dx=x\sin^{-1}(ax)-\int x\frac{d}{dx}\sin^{-1}(ax) dx$$Looking up or working out the deritative of $\sin-1(ax)$ produces $\displaystyle I=\sin^{-1}(ax)-\int\frac{ax}{\sqrt{1-a^2x^2}}dx$
The integral now has the form of the numerator (top) being related to the derivative of the denominator, equation 12, and the final result is
$$\displaystyle I=\sin^{-1}(ax) +\frac{\sqrt{1-a^2x^2}}{a}$$(a) Changing the limits is described by equation 5, the general integral becomes
$$\displaystyle \int_a^b f(x)dx=g(b)-g(a)=-\int_b^a f(x)dx$$by swapping limits and
$$\displaystyle \int_{-a}^{-b} f(x)dx=g(-b)-g(-a)=\int_a^b f(-x)dx$$by negating limits.
In this particular example, negating the limit gives $\displaystyle Ei(-x)=-\int_x^\infty \frac{e^{-t}}{t}dt$
and swapping limits $\displaystyle Ei(-x)=-\int_\infty^x \frac{e^{-t}}{t}dt= \int_{-\infty}^x\frac{e^t}{t}dt$.
and the last step follows from the properties of limits, see Section 1.3.
(b) Integration by parts gives
$$\displaystyle -Ei(x)= \int_{-x}^\infty \frac{e^{-t}}{t}dt=-\frac{e^{-t}}{t}\bigg|_{-x}^\infty - \int_{-x}^\infty \frac{e^{-t}}{t^2}dt = \frac{e^{x}}{x} - \int_{-x}^\infty \frac{e^{-t}}{t^2}dt $$And repeating the process the next integral is
$$\displaystyle -\int_{-x}^\infty \frac{e^{-t}}{t^2}dt=\frac{e^{-t}}{t^2}\bigg|_{-x}^\infty +2 \int_{-x}^\infty \frac{e^{-t}}{t^3}dt = -\frac{e^{x}}{x^2} +2 \int_{-x}^\infty \frac{e^{-t}}{t^3}dt $$Repeating the process $n$ times a pattern becomes clear and is
$$\displaystyle \int_{-x}^\infty \frac{e^{-t}}{t}dt=-\frac{e^{-t}}{t^n}\bigg|_{-x}^\infty -n\int_{-x}^\infty \frac{e^{-t}}{t^{n+1}}dt$$and adding these integrals gives
$$\displaystyle -Ei(x)\approx \frac{e^x}{x} - \frac{e^x}{x^2} + 2\frac{e^x}{x^3}-6\frac{e^x}{x^4}+24\frac{e^x}{x^5} - \cdots (-1)^n n!\frac{e^x}{x^{n+1}} - \cdots n!\int_{-x}^\infty \frac{e^{-t}}{t^{n+1}} $$and the summation can be expressed as
$$\displaystyle Ei(x)\approx e^x\sum_{n=0} (-1)^n\frac{n!}{x^{n+1}}$$Calculating this summation is a frustrating affair, the number produced are typically very large for example $\approx 10^{20}$ when $x=50$. The results are comparable to the answer Python/Scipy gives for large $x$, say, $\pm 500$ and for $30$ terms in the summation. However, with $x \gt \approx 50$, the results are unpredictable and vary with the number of terms in the summation. Clearly this summation is unstable and not a good way of evaluating this integral.
The instability for small $x$ arises because $\pm n!/x^{n+1}$ becomes larger and larger as $n$ increases and the sign alternates. This result shows how an algebraic formula, while correct, can give poor answers in a numerical calculation because of rounding errors; see Chapter 11. This is particularly true of summations with alternating positive and negative terms as in this case. Python/scipy uses the built-in function $\mathtt{expi(x)}$ to numerically evaluate this integral. (You need to add 'from scipy.special import expi' to import this function into a python worksheet).
(a) Recognize this as an integration by parts and use equation 16. If $u = x$ then $dv = e^{ax}dx$ and the formula gives
$$\displaystyle \int xe^{ax}dx =x\frac{e^{ax}}{a}-\frac{1}{a}\int e^{ax} dx=\frac{(ax-1)}{a^2}e^{ax}$$(b) The result of the previous calculation produces
$$I_1= xI_0 - I_0/2\quad\text{ where }\quad I_0=\int e^{ax}dx=e^{ax}/a$$Calculate the next integral by letting $u = x$ then $dv = e^{ax}dx$ and the integration by parts formula gives
$$\displaystyle \int x^2 e^{ax}dx=\frac{x^2}{a}e^{ax}-\frac{2}{a}\int xe^{ax}dx$$which can be written as
$$\displaystyle I_2=x^2I_0-\frac{2}{a}I_1 \tag{97}$$using $I$ from the question and where, logically, $I_0$ is defined as $\displaystyle I_0=\int e^{ax}dx = a^{ax}/a$. Next, define $\displaystyle I_3=\int x^3e^{ax}dx$ which is $I_3=x^2I_0-3i_2/a$ and then by induction
$$\displaystyle I_n=x^nI_0-\frac{n}{a}I_{n-1} \tag{98}$$which can also be written as a recursion formula, $aI_n+nI_{n-1}=x^ne^{ax}$. The code below shows the recursion using Sympy. This gives the same answer as direct integration.
a, x = symbols('a, x',positive = True)
def Intn(n):
if n == 0: return exp(a*x)/a
if n == 1: return (a*x - 1)*exp(a*x)/a**2
return (x**n*exp(a*x) - n*Intn(n - 1))/a # recursion formula
#-----------------
simplify(Intn(5) ) # integral x^5 exp(a*x)
(a) The integral for $n=1$ can be calculated between $0$ and infinity but the way to do this is a little cunning. The normal 'by parts' integral is $\displaystyle \int u dv = uv - \int vdu$. Now, suppose that $\displaystyle v=e^{-ax^2}$ therefore $dv=-2axe^{-ax^2}$ and also if $u=1$ then
$$\displaystyle I_1=\frac{1}{2a}\int_0^\infty 1(-2axe^{-ax^2})dx$$which can now be integrated by parts. The $\displaystyle -\int vdu$ integral on the right is zero because $u = 1$. The result is
$$\displaystyle I_1=-\frac{e^{-ax^2}}{2a}\bigg|_0^\infty -\int0dx=\frac{1}{2a}$$(b) Letting $v = x^n$ because $e-x^2$ cannot be integrated easily, gives
$$\displaystyle I_n=e^{-ax^2}\frac{x^{n+1}}{n+1}\bigg|_0^\infty + 2a\int_0^\infty xe^{-ax^2}\frac{x^{n+1}}{n+1}dx$$which is
$$\displaystyle I_n=e^{-ax^2}\frac{x^{n+1}}{n+1}\bigg|_0^\infty + \frac{2a}{n+1}I_{n+2}$$To work out the limits, when $x = 0$, then so is $e^{ax^2}x^{n+1}$. When $x = \infty$, repeated differentiation using l'Hopital's rule produces
$$\displaystyle \lim_{x\to\infty} \frac{x^{n+1}}{e^{ax^2}(n+1)}\to \frac{x^{n-1}}{2ae^{ax^2}}\to\to 0$$because for any $n$ the numerator will become eventually become 1, while the denominator is still infinity thus making the limit zero. Since both limits are zero then $\displaystyle I_{n+2}=\frac{n+1}{2a}I_n$ which is more conveniently re-written as $\displaystyle I_{n}=\frac{n-1}{2a}I_{n-2}$ for calculation.
The values of all other integrals of this type can now be calculated provided that $I_0$ is known and this has the value $I_0 = \pi/4a$.
a, x = symbols('a, x',positive =True)
def intxn(n):
if n == 0: return sqrt(pi/(4*a) )
if n == 1: return 1/(2*a)
if n == 2: return sqrt(pi/a)/(4*a)
return (n - 1)*intxn(n - 2)/(2*a) # recursion chnage n to n-2
#-----------------
ans = []
for n in range( 6): # make list to see results
ans.append(n)
ans.append(intxn(n) )
ans
The strategy here is to recognize that $d\ln(x)=1/x$ or, equivalently, letting $w=\ln(x)$ (and $dw=dx/x$) the integral becomes
$$\displaystyle \int \frac{\cos(\ln(x))}{x}dx= \int \cos(\ln(x)) d\ln(x)\equiv\int\cos(w)dw=\sin(\ln(x))+c$$The secant function is the reciprocal of cosine and its integral is a standard one, $\displaystyle \int \sec^2(x)dx = \tan(x)$. The square root is removed by substitution, therefore try $z=\sqrt{x}$ and $dz/dx=1/\sqrt{4x}=1/2z$ making the integral,
$$\displaystyle \int\sec^2(\sqrt{x})dx=2\int z\sec^2(z)dz$$This integral can now be tackled by the 'parts' method,
$$\displaystyle 2\int z\sec^2(z)dz=2z\tan(z)-2\int \tan(z)dz=2z\tan(z)+2\ln(\cos(z)) +c$$where $c$ is a constant of integration. Substitute back for $z$ to get the result.
(a) This integral can be evaluated by converting to the exponential form but can also be done rather easily by parts;
$$\displaystyle \int \cosh(x)\sinh(x)dx=\sinh^2(x)-\int \cosh(x)\sinh(x)dx$$and rearranging gives $\displaystyle \int\cosh(x)\sinh(x)=\frac{1}{2}\sinh^2(x)+c$
(b) Using the example in section 5 to evaluate the integrals gives $\displaystyle \int x\cos(x)dx=\frac{e^x}{2}\left( \sin(x)+\cos(x) \right)$
Converting to an exponential form gives
$$\displaystyle I= \int \frac{dx}{\sin(ax)}=2i\int \frac{dx}{e^{iax}-e^{-iax}}$$which appears to have no easy solution. Try substituting $z=e^{iax},\; dz=iae^{iax}dx$ giving
$$\displaystyle I=\frac{2}{a}\int \frac{dz}{z^2-1} $$which has a standard form, (see section 2.13) which is $\displaystyle I= \frac{1}{2}\ln\left(\frac{z-1}{z+1} \right)$ making the integration
$$\displaystyle \int \frac{dx}{\sin(ax)}=\frac{1}{2a}\ln\left(\frac{\cos(ax)-1}{\cos(ax)+1} \right)+c$$Some texts may convert the answer into hyperbolic form using $\displaystyle \tanh^{-1}(z)=\frac{1}{2}\ln\left( \frac{1+z}{1-z} \right)$ where $z$ is defined above. In this case the result has to be real even though $z$ is complex, and even though the result may appear to be complex, the area under the curve, which is the integral, cannot be complex since $1/\sin(ax)$ is real.
(a) $\displaystyle \int_{-\pi}^\pi dx=x\bigg|_{-\pi}^\pi =2\pi$.
(b) Integrating and then changing the exponential to a sine using $2i\sin(x)=e^{ix}-e^{-ix}$ gives
$$\displaystyle \int_{-\pi}^\pi e^{imx}e^{-inx}dx =-\frac{e^{-ix(m+n)}}{i(m+n)}\bigg|_{-\pi}^\pi=-\frac{ e^{-i\pi(m+n) } }{i(m+n) }+\frac{e^{i\pi(m+n)}}{i(m+n)}=2\frac{\sin\left( n\pi+m\pi \right)}{m+n}$$Recalling that $m$ and $n$ are integers let $q = n + m$, which can be positive, negative, or zero. Consider the case first when $q \ne 0$ then the integral is zero because $\sin(\pi q) = 0$ for any integer value of $q$. In the case that $q = 0$ then the denominator is zero and using l'Hopital's rule (chapter 3.8) the limit is
$$\lim_{q\to 0} 2\frac{\sin(q\pi}{q}\to 2\frac{\pi\cos(q\pi)}{1}\to 2\pi$$Together these results show that $\displaystyle \int_{-\pi}^\pi e^{imx}e^{-inx}dx=2\pi\delta_{n,m}$.
Differentiating with respect to $\beta$ only involves the exponential and produces
$$\displaystyle \frac{d}{d\beta}\int_0^\infty \frac{e^{-\beta x}\sin(x)}{x}dx=\int_0^\infty e^{-\beta x}\sin(x)dx$$Integrating the result by parts produces
$$\displaystyle \int_0^\infty e^{-\beta x}\sin(x)dx =-e^{-\beta x}\cos(x)\bigg|_0^\infty -\beta\int_0^\infty e^{-\beta x}\cos(x)dx $$Continuing the integration as in Section 5, example (c) and evaluating gives $\displaystyle \int_0^\infty e^{-\beta x}\sin(x)dx=\frac{1}{1+\beta^2}$. This result has to be integrated again to obtain the final result but now with respect to $\beta$ which is the standard integral $\displaystyle \int\frac{d\beta }{1+\beta^2 }=\tan^{-1}(\beta)\bigg|_0^\infty =\frac{\pi}{2}$. The final result is
$$\displaystyle \int_0^\infty \frac{\sin(x)}{x}dx = \frac{\pi}{2}$$Using the definition of the cosine in a right-angled triangle, and in the limit of small values,
$$\displaystyle \cos(x)=\frac{dx}{\sqrt{dx^2+dy^2}}=\left(1+\left(\frac{dy}{dx} \right)^2 \right)^{-1/2} $$then using the equation in the question $\displaystyle 2\pi yT \left(1+\left(\frac{dy}{dx} \right)^2 \right)^{-1/2} =c$ and rearranging to give
$$\displaystyle \frac{dy}{dx}=\sqrt{\left( \frac{2\pi T}{c} \right)^2y^2-1}= \sqrt{c_0^2y^2-1}$$and, for clarity, the constants are replaced by $c_0$. This equation is now integrated by separating $dy$ and $dx$
$$\displaystyle \int \frac{dy}{\sqrt{c_o^2y^2-1}}=\int dx=x+const$$The integral in $y$ evaluates to $\displaystyle \int \frac{dy}{\sqrt{c_o^2y^2-1}}=\frac{1}{c_0}\cosh^{-1}(c_0y)$ which produces
$$\displaystyle y=\frac{\cosh(c_0x)}{c_0}$$The constants can be determined by defining the geometry of the soap film; the equation used is defined with $y = r$ at $x = 0$ therefore, $c_0 = 1/r$ because $\cosh(0)=1$