When doing substitution problems in integration, I've often come across things like $\mathrm{d}x/\mathrm{d}\theta =\cos \theta$ and so $\mathrm{d}x=\cos\theta \mathrm{d}\theta$, which seems like $\mathrm{d}x$ divided by $\mathrm{d}\theta$? I've asked it in a comment, by I got a complicated answer which I can't understand as I am a newbie in calculus. So could anyone give me a simple and step by step answer to explain why is that the case?
14091
I just finished taking my first year of calculus in college and I passed with an A.  I don't think, however, that I ever really understood the entire $\frac{dy}{dx}$ notation (so I just focused on using $u'$), and now that I'm going to be starting calculus 2 and learning newer integration techniques, I feel that understanding the differential in an integral is important. Take this problem for an example: $$ \int 2x (x^2+4)^{100}dx $$ So solving this... $$u = x^2 + 4 \implies du = 2x dx \longleftarrow \text{why $dx$ here?}$$ And so now I'd have: $\displaystyle \int (u)^{100}du$ which is $\displaystyle \frac{u^{101}}{101} + C$ and then I'd just substitute back in my $u$. I'm so confused by all of this.  I know how to do it from practice, but I don't understand what is really happening here.  What happens to the $du$ between rewriting the integral and taking the anti-derivative?  Why is writing the $dx$ so important?  How should I be viewing this in when seeing an integral?
102336
Given  $$x=r\,\cos\theta\\y=r\,\sin\theta$$ and $$x'=r'\,\cos\theta'\\y'=r'\,\sin\theta'$$  how can I express  $$\delta(x'-x)\delta(y'-y)$$ in terms of the polar coordinates?  And the more general case:  $$\delta(x'-x-a)\delta(y'-y-b)$$
110064
Here is a problem from Griffith's book Introduction to E&amp;M.     Let $\theta(x)$ be the step function   $$\theta = \begin{cases} 0, &amp; x \le 0, \\ 1, &amp; x \gt 0.  \end{cases} $$      The question is how to prove $\frac{d\theta}{dx} = \delta(x)$.   I think since the function is discontinuous at $x = 0$, there is no definition of $\frac{d\theta}{dx}$ at the point $x = 0$ at all. Thus, how could we show the equivalence of a equation if the left part of equation is not defined at the point $x = 0$?
37269
This question: got some attention, but I didn't understand anything of it. Now I looked at the solution of the book and there is given the hint to use the change of variable $$x=\frac{1-t}{1+2t}$$ Could you please explain how one could think of such substitution? Is there some intuition behind this? Also after I used the substitution I got: $$\int_{-1/5}^1\frac{3\arctan\left(\frac{1-t}{1+2t}\right)}{5t^2+8t+5}dt$$ How can I finish this using the hint from the book?
5258
I see an equation like this:  $$y\frac{\textrm{d}y}{\textrm{d}x} = e^x$$  and solve it by "separating variables" like this:  $$y\textrm{d}y = e^x\textrm{d}x$$ $$\int y\textrm{d}y = \int e^x\textrm{d}x$$ $$y^2/2  = e^x + c$$  What am I doing when I solve an equation this way?  Because $\textrm{d}y/\textrm{d}x$ actually means  $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$$  they are not really separate entities I can multiply around algebraically.    I can check the solution when I'm done this procedure, and I've never run into problems with it.  Nonetheless, what is the justification behind it?  What I thought of to do in this particular case is write  $$\int y \frac{\textrm{d}y}{\textrm{d}x}\textrm{d}x = \int e^x\textrm{d}x$$ $$\int \frac{\textrm{d}}{\textrm{d}x}(y^2/2)\textrm{d}x = e^x + c$$  then by the fundamental theorem of calculus  $$y^2/2 = e^x + c$$  Is this correct?  Will such a procedure work every time I can find a way to separate variables?
54135
The integral to be solved is $\int \sqrt{x^8 + 2 + x^{-8}} \,\mathrm{d}x$. I tried substitution $t=x^8$ which got me to $\frac{1}{8} \int \sqrt{t + 2 + \frac{1}{t}} t^{-7/8} \,\mathrm{d}t$ and I'm stuck. Can you help?
114861
Is this correct? I thought it would be but when I entered it into wolfram alpha, I got a different answer.   $$\int (\cos^3x)(\sin^2x)dx  = \int(\cos x)(\cos^2x)(\sin^2x)dx  = \int (\cos x)(1-\sin^2x)(\sin^2x)dx.$$  let $u = \sin x$, $du = \cos xdx$  $$\int(1-u^2)u^2du = \int(u^2-u^4)du  = \frac{u^3}{3} - \frac{u^5}{5} +C$$  Plugging in back $u$, we get $\displaystyle\frac{\sin^3 x}{3} - \frac{\sin^5 x}{5}$ + C
34118
$$\int_{0}^{\frac{\pi }{2}}x\cot(x)dx$$   I tried integration by parts and got $\dfrac{1}{2}\int_{0}^{\frac{\pi }{2}}x^{2} \csc^{2}x dx$ which doesn't help at all. I don't really know what to do. Any help will be greatly appreciated.
13331
I've tried substituting $1+\cos\theta=2\cos^2(\frac{\theta}{2})$ but it didn't give me the answer.
69861
$$\int{\frac{1+x^2}{(1-x^2)\sqrt{1+x^4}}}\mathrm dx$$   This was a Calc 2 problem for extra credit (we have done hyperbolic trig functions too, if that helps) and I didn't get it (don't think anyone did) -- how would you go about it?
11731
How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$... Now I'm stuck. I don't know how to proceed using substitution rule.
118918
Find $\displaystyle \int \dfrac{x^3}{x^2-1} \:\mathrm{d}x$.   I think there are a few ways to do this. We could just break this up using long division or break it up into partial fractions then use long division. My homework says to break this up into its roots so that's why I brought up partial fractions. Is that the best way to do it here?
104646
How would I carry out the following integration? $$\int_0^12\arctan x^2$$  I tried to substitute $x^2 = \tan\theta$ From there I wrote the integral as: $$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$  Now, is my only option to use integration by parts? Or is there a better method to solve this integral?
68815
I have to evaluate:  $$\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x. $$ I can't get the right answer! So please help me out!
117001
Evaluate $$\int \frac{2-3x}{2+3x}\sqrt\frac{1+x}{1-x}d x$$ What substitution should I use ? $\sqrt\frac{1+x}{1-x}$ suggests $x=cos2\theta $ but its not useful in $\frac{2-3x}{2+3x}$
90897
Integration by parts - unless I'm not thinking straight - doesn't seem to help here.. If I pick either of the functions $e^{2x}$ or $\sin(3x)$ to be $u$ or $dv$, they don't change to anything easier... $e^{2x}$ stays in the form $e^x$ and $\sin(3x)$ flip-flops between $\sin(3x)$ and $\cos(3x)$, neglecting the constant that gets introduced.  So how can I approach solving this? $u$-substitution doesn't seem to be something I can use either.
14121
It is changing the coordinate from one coordinate to another. There is an angle and radius on the right side. What is it? And why?  I got:  $2\,\mathrm dy\,\mathrm dx = r(\cos^2\alpha-\sin^2\alpha)\,\mathrm d\alpha \,\mathrm dr$,  where $x = r \cos(\alpha)$ and $y = r \sin(\alpha)$.  but cannot understand and get the right side. The problem emerged when trying to integrate $\displaystyle \int_0^\infty e^{\frac{-x^2}{n^2}}\,\mathrm dx$ where I tried to change the problem knowing $r^2=x^2+y^2$ but stuck to this part. What is the change in the title called and why is it so?