When using the subsitituion rule in integration of an integral $\displaystyle \int f(x)\,\mathrm dx$, one turns the integral from the form $$\displaystyle \int f(x(t))\,\mathrm dx \quad(*)$$ into the form $$\int f(x(t))\,\frac{\mathrm dx}{\mathrm dt}\mathrm dt \quad(**)$$. This transform is usually accomplished by means of differenting the subsitution $x = x(t)$, such that $\dfrac{\mathrm dx}{\mathrm dt} = \dfrac{\mathrm d x(t)}{\mathrm dt}$. Now, at this point, one turns this into a differential form by means of magic, s.t. $\mathrm dx = \dfrac{\mathrm dx(t)}{\mathrm dt}\mathrm dt$. This now substitutes the differential term $\mathrm dx$ in the original expression $(*)$ to the one in the transformed expression $(**)$.  I'd like to learn that magic step a bit more rigorous – so that I can better understand it. It is often explained by "multiplication" of $\mathrm dt$, which do make sense, but it does not explain the nature of differentials; when is "multiplication" allowed? It seems there should be a more rigorous way of explaining it, perhaps by defining the "multiplication.  So, in what ways can differentials like $\mathrm dx$ and $\mathrm dt$ be formalized in this context? I've seen them being compared to small numbers, which often work, but can this analogy fail? (And what are the prerequisites needed to understand them?)
7365
In the book Thomas's Calculus (11th edition) it is mentioned (Section 3.8 pg 225) that the derivative $dy/dx$ is not a ratio. Couldn't it be interpreted as a ratio, because according to the formula $dy = f'(x)dx$ we are able to plug in values for $dx$ and calculate a $dy$ (differential). Then if we rearrange we get $dy/dx$ which could be seen as a ratio.   I wonder if the author says this because $dx$ is an independent variable, and $dy$ is a dependent variable, for $dy/dx$ to be a ratio both variables need to be independent.. maybe?
66971
I am told that an expression like $$ \int_a^x f(x)dx $$ is not well formed, i.e. it should be $$ \int_a^xf(t)dt $$ or similar.  Why is it that the limits of integration can't depend on the variable of integration?
79451
I would like to understand why the integrals represent 2 things:   If we only do an integral without bounds on $f(x)$, the integral gives us the function $F(x)$ that, when differentiated, gives us the original function $f(x)$. If we do an integral with bounds $[a,b]$ on $f(x)$, it gives us the area under $f(x)$ by evaluating the difference $F(b) - F(a)$.   Why does $F(x)$ represent the area under $f(x)$ as well as the first point?
76330
I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique.   When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely.   But I remember that the technique I saw was a nice way of evaluating these even when $a,b\neq 1$.
14940
$$\int \frac{\mathrm{d}x}{x}$$ If I integrate this by parts ($\displaystyle u=\frac{1}{x}, \mathrm{d}u = -\frac{\mathrm{d}x}{x^2}, \mathrm{d}v= \mathrm{d}x, v = x$), then why does this happen?  $$\int \frac{1}{x} \: \mathrm{d}x = u v - \int v \:\mathrm{d}u = \frac{x}{x} - \int x\left(\frac{-1}{x^2}\right) \:\mathrm{d}x=1+\int \frac{1}{x}\:\mathrm{d}x$$  So $\displaystyle\int \frac{1}{x}\:\mathrm{d}x = 1 + \int \frac{1}{x} \:\mathrm{d}x$ and cancel out both sides of $\displaystyle\int \frac{1}{x} \:\mathrm{d}x$ and I get $0=1$. How can I find the error in the process because right now I cant see it.
85114
I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.  Any suggestions are welcome. Thanks.
40411
So far I've done this, but I don't know if it will help.  Let $\int_0^xf(t)dt=F(x)$.  Now, $1=\int xf(x)=F(x)x|_0^1-\int_0^1F(x)dx=1-\int_0^1 F(x)dx$. Then, $\int_0^1 F(x)dx=0$.  But I don't know how to keep going.
83211
What the following formula mean?  $$\int\mathop{}\!\mathrm{d}^4x$$  I know that this $\int f(x)\mathop{}\!\mathrm{d}x$ is the integral of the function $f$ over the $x$ variable, but the following $\int\mathop{}\!\mathrm{d}^4x$ leave empty the argument and also have the $\mathrm{d}$ elevated to the fourth power. What that mean?  Update.  In the Einstein-Hilbert action we have (note that I have understand that the other parts of the integral are relevant only after your answers, thus sorry.):  $$S=\frac{c^4}{16\pi G}\int\mathop{}\!\mathrm{d}^4x \, \ R \sqrt{-g}$$
87562
I've noticed that in QFT literature, integrals are usually written as $\int \!dx ~f(x)$ instead of $\int f(x) dx$. Why?
47708
I am looking to find out how the second derivative formula works.  I can blindly apply it but I don't have a grasp of what is going on or why. The first derivitive has the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. I told this was because if you divide them you cancel out the dts and get dy/dx. The second derivative has the formula $\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{ \frac{d}{dt}\left(\frac{dy}{dx}\right) }{\frac{dx}{dt}}$. Is this something that can be explained with only a knowledge of calculus or is this something I would have to take on faith until I learn more. Any help would be greatly appreciated.
43856
How do I solve an integral with a differential on top? E.g.: given this integral to evaluate: $$\;\int \frac{dt}{(\cos(t))^2}\;\;?$$  What does it even mean when there's a differential?
117830
How to integrate $\frac{1}{x\log x}$?  Could you give me some ideas on how to integrate this? thanks.  i've tried setting $u=(\log x)^{-1}$.   $\dfrac{\mathrm du}{\mathrm dx} = x^{-1}$  But it didnt work...
114165
$$\int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$   $\bf{My\; Try::}$ Let $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\frac{1}{e}}^{\cot x}\frac{1}{t(1+t^2)}dt$$  Put $\displaystyle x = \frac{1}{u},$ Then $\displaystyle dx = -\frac{1}{u^2}du$ and changing limits, We get  $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{1}{u(1+u^2)}du$$  So $$f(x) = \int_{\frac{1}{e}}^{\tan x}\frac{t}{1+t^2}dt+\int_{\tan x}^{e}\frac{t}{1+t^2}dt = \frac{1}{2}\int_{\frac{1}{e}}^{e}\frac{2t}{1+t^2}dt$$  So $$f(x) = \frac{1}{2}\ln (1+t^2)|_{\frac{1}{e}}^{e} = \ln(e) = 1$$  My question is how can we solve it in some shorter way, Help required, Thanks
3043
I want a hint on this following integral. I don't know the first step, do I do $u$-sub. or do I use trig. sub or something?  $$\int \frac{1}{2+ \sin x + \cos x} ~\mathrm{d}x$$
113522
Please refer to my school textbook pg48 (of the book, and not the pdf counter) here:   My doubt is in this context: (right side column)  $a = dv/dt = v dv/dx$  then integrating $v$ with respect to $dv$, and $a$ with respect to $dx$.  Now, when we integrate $a$, either we say it is a constant, and give $a\times(x-x_o)$ as a result of the integration, or we say it is not constant, and is a function of time, in which case, it can not be integrated like this. But the book integrates it like this, and then it is written:     The advantage of this method is that it can be used for motion with non-uniform acceleration also.   Can you please explain how it can be used with non uniform acceleration while it has been assumed while integrating that $a$ is constant? And moreover, does this line seem to apply to only this derivation, or does it apply to other two before it as well?
95678
could you help me on this integral ? $$\int \frac{1}{\cos(x)}\,\mathrm dx$$ Here's what I've started : $$\int \frac{1}{\cos(x)}\,\mathrm dx = \int \frac{\cos(x)}{\cos(x)^2}\,\mathrm dx = \int \frac{\cos(x)}{1-\sin(x)^2}\,\mathrm dx$$ Now, I did : $u = \sin(x)$, so $\mathrm du = 1$. Now I have : $$\int \frac{\text{???}}{1-u^2}\,\mathrm du$$ But at this point, I think I did the most of the job but I'm stuck. Could you help me to solve this integral please (to the integration by substitution at the end) ? Thanks EDIT : Now I follow the steps and I got : $$\int \frac{1}{1-u^2}\,\mathrm du$$ Doing the partial fraction I got $A = 1/2$ and $B = 1/2$. So basically I have : \begin{align} &amp; \int \frac{1}{1-u^2}\,\mathrm du = \int \frac{1/2}{1+u}\,\mathrm du + \int \frac{1/2}{1-u}\,\mathrm du \\[8pt] = {} &amp; \frac 1 2 \left(\int \frac{1}{1+u} \, du - \int \frac{1}{1-u} \, du\right) \\[8pt] = {} &amp; \frac 1 2 \ln\left(\frac{1+u}{1-u}\right) = \ln\left(\left(\frac{1+\sin(x)}{1-\sin(x)}\right)^{1/2}\right) \\[8pt] = {} &amp; \ln \left(\frac{\sqrt{1+\sin(x)}}{\sqrt{1-\sin(x)}}\right) = \ln\left(\frac{\sqrt{1+\sin(x)}}{\sqrt{\cos(x)^2}}\right) \\[8pt] = {} &amp; \ln\left(\frac{\sqrt{1+\sin(x)}}{\cos(x)}\right) \end{align} Here's what my teacher got :  What's wrong with what I did ? Did I miss something ?
6506
$$\int \frac{1}{(x^2+1)^2}\mathrm dx$$ Look simple, but I stuck a little bit. What method is better in this situation? Please, help.