I am confused why we can introduce differentials into an integral when performing an integration by substitution.   Consider the integral $$\int \frac{1}{ x \sqrt{1-x} } dx.$$ We can perform the substitutions $$x=\sin^2u,$$ $$dx=2\sin u \cos{u} du  ,$$ on the integral to give $$\int \frac{1}{ \sin^2u \sqrt{1-\sin^2u} } 2\sin u \cos u du .$$  Why can you treat $dx$ as a differential?   From my understanding the integration sign $\int dx$ works as as if it is an operator, just like how $\frac {d}{dx}$ works as an operator and not as a fraction. Which means $\int$ and $dx$ should not be interpreted separately.   But at the same time treating $dx$ as a differential always work out fine so there must be some validness in treating it as an differential.
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?
100415
$$\int\frac{(\ln x)^{10}}{x}\,dx$$ All I know is that I am supposed to substitute $u=\ln x$. But can someone please explain to me how to find the anti derivative of $(\ln x)^{10}$. I think we are supposed to use integration by parts on integrating $\ln x$, but we haven't been taught that yet.
104093
$$\int{\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx}$$  $$\int\sqrt{\frac{(1-\sin x)(2-\sin x)}{(1+\sin x)(2+\sin x)}}dx=\int \frac{(1-\sin x)(2-\sin x)}{\sqrt{(1-\sin x)(2-\sin x)(1+\sin x)(2+\sin x)}}dx$$  I am stuck. Please help me....
88435
Evaluate the indefinite integral   $$\int \frac{x^2 \,dx}{a x^5 + b},$$   for real parameters $a, b \neq 0$.   No apparent substitutions simplify the expression (if the exponent of $x$ were an integral multiple of $3$, this would suggest the substitution $u = x^3$, $du = 3 x^2 \,dx$, but the exponent is not). Applying integration by parts with $u = a x^5 + b$, $dv = x^2 dx$ is straightforward, but it produces an integrand with a much larger degree in the denominator and so appears only to make the situation worse. Applying integration by parts instead with $dw = \frac{x^k dx}{a x^5 + b}$ results in integrating $\frac{x^k dx}{a x^5 + b}$, which except when $k \equiv 4 \pmod 5$ (which doesn't appear immediately fruitful), doesn't appear much easier than the given integral.
122986
Quick question, might end up having a simple answer, but I have here a "proof" that any integral from 0 to $2\pi$ is zero, as follows:  $$\int^{2\pi}_0f(x)dx$$  Now using u-substitution, let $u = \sin x$, so $dx = \frac{du}{\cos x}$, so:  $$\int^{0}_{0}\frac{f(x)du}{\cos x}$$  (The bounds were evaluated to zero due to the u-substituion).  However, this is zero, since the integral goes from zero to zero.  Any help as to why this is wrong would be appreciated.
69250
$\int_{0}^{1}\frac{\sin^{-1}\sqrt x}{x^2-x+1}dx$  Put $x=\sin^2\theta,dx=2\sin \theta \cos \theta d\theta$  $\int_{0}^{\pi/2}\frac{\theta.2\sin \theta \cos \theta d\theta}{\sin^4\theta-\sin^2\theta+1}$ but this seems not integrable.Is this a wrong way?Is there a different way to solve it?
104018
If we have to integrate $$f(x)=\frac{x}{\sqrt{1-x^2}}$$ and we substitute $x=\sin \theta$ then we eventually have to take the square root of $\cos^2x$ which is equal to $|\cos x|$.  But in my textbook and class lectures, we simply remove the absolute value sign and replace it with $\cos x$.  Why do we do this? I don't understand how we can do this.
9574
Integrate  $\displaystyle\int\dfrac{2x^2}{2x\cos(2x)+(x^2-1)\sin(2x)} \mathrm d x$   I tried dividing by $\cos^2(x)$ and then substituting $\tan(x)=t$.
95025
Find the indefinite integral  $$\int {dx \over {(1+x^2) \sqrt{1-x^2}}} $$  Is there a smart substitution or algebric trick that I'm missing? Because integration by parts hasn't helped..
93963
I don't know how to integrate $\int x^2\sin^2(x)\,\mathrm dx$. I know that I should do it by parts, where I have  $$ u=x^2\quad  v'=\sin^2x \\  u'=2x \quad v={-\sin x\cos x+x\over 2}$$  and now I have  $$ \int x^2\sin^2(x)\,\mathrm dx = {-\sin x\cos x+x\over 2} x^2 - \int 2x{-\sin x\cos x+x\over 2}\,\mathrm dx\\ ={-\sin x\cos x+x\over 2} x^2 - \int x(-\sin x\cos x+x)\,\mathrm dx\\  $$  so I have to calculate $$ \int x(-\sin x\cos x+x)\,\mathrm dx=-\int x\sin x\cos x\,\mathrm dx+\int x^2\,\mathrm dx$$  I know that $\int x^2\,\mathrm dx = {1 \over 3}x^3+C$ but I don't know what should I do with $$\int x\sin x\cos x \,\mathrm dx$$ Should i use parts again or what? Please help.
97918
Integration Of a given function can be found out in many ways,  For a specific function ∫1/xlogx, if we do integration by parts (∫f(x) g(x)= f(x) ∫ g(x)- ∫ [d/dx (f(x)) ∫g(x)] dx ) we get this way  Let ∫1/xlogxdx = A  If we do integration by parts, we have  $$\int \frac{1}{x\log x} \, dx= \frac{1}{\log x}\cdot\int \frac{1}{x} \, dx - \int \frac{d}{dx}\left(\frac{1}{\log x}\right)\,dx\cdot\int \frac{1}{x} \, dx+c$$ $$=\frac{1}{\log x}\cdot \log x+\int \frac{1}{x\log x}\,dx+c$$  i.e, A = 1 + A + c   Which tells us that the value of "$c$" in this specific integral is $-1$.  So does this mean that the Integral of function $\dfrac{1}{x\log x}$ has only one value.! So this means only two things, either I am wrong somewhere or i am  missing a point somewhere.   Edit: I do know we can simple get it by writing logx as u and ¹/x as du, we directly get it as log(logx) +c.  But I wanted to know the reason why we can't apply By parts when we can integrate it by substitution.
59734
$$ \int \sin \theta\cos \theta~d \theta= \int \frac {1} 2  \sin 2\theta~ d \theta=-\frac {1} 4  \cos 2\theta$$ But, if I let  $$ u=\sin \theta , \text{ then }du=\cos \theta~d\theta $$ Then $$ \int \sin \theta\cos \theta~d \theta= \int u ~ du =\frac { u^2 } 2 =\frac {1} 2 \sin^2 \theta $$ Since $$ \sin^2 \theta =\frac {1} 2 - \frac {1} 2 \cos 2\theta$$ The above can be written as $$ \int \sin \theta\cos \theta~d \theta= \frac {1} 2 \sin^2 \theta =\frac {1} 2 \left( \frac {1} 2 - \frac {1} 2 \cos 2\theta \right)=\frac {1} 4-\frac {1} 4 \cos 2\theta   $$ Why are the two results differ by the constant $1/4$? Thank you.
39315
Can the order of a differentiation and summation be interchanged, and if so, what is the basis of the justification for this? E.g. is $\frac{\mathrm{d}}{\mathrm{d}x}\sum_{n=1}^{\infty}f_n(x)$ equal to $\sum_{n=1}^{\infty}\frac{\mathrm{d}}{\mathrm{d}x}f_n(x)$ and how can it be proven? My intuition for this is that it should be the case, since in the limit, the summation becomes an integral and this can be interchanged with the differentiation operator, but I don't know how to justify this? Would it suffice to say that $\frac{\mathrm{d}}{\mathrm{d}x}f_1(x)+\frac{\mathrm{d}}{\mathrm{d}x}f_2(x) = \frac{\mathrm{d}}{\mathrm{d}x}(f_1(x)+f_2(x)$ because it is a linear operator and then somehow extend this to an infinite sum? Thanks in advance!
64878
Here is a toned down example of what I'm looking for:  Integration by solving for the unknown integral of $f(x)=x$:  $$\int x \, dx=x^2-\int x \, dx$$  $$2\int x \, dx=x^2$$  $$\int x \, dx=\frac{x^2}{2}$$  Can anyone think of any more examples?  P.S. This question was inspired by a  that I found out about  This question is meant to be more general, accepting things like solving integrals and using complex numbers to evaluate simple problems.
89670
Today I read the answer to , in which the poster integrates $x^5e^x$ by making these manipulations with the differential operator $D$: $$\frac1Dx^5e^x=e^x\frac{1}{1+D}x^5=e^x(1-D+D^2+...)x^5$$  which I was amazed by but yet suspicious of. After reading a bit on differential operators, I know a few properties. For example, $D+a$ (where $a$ is constant) is a polynomial differential operator which comes from the differential equation $y'+ay = q(x)$. Also, for polynomials $f$ and $g$, $f(x)\cdot g(x)=h(x)$ implies $h(D)u=f(D) \circ [g(D)u]$ with $u$ being a function and $f(D), g(D), h(D)$ being polynomial differential operators. Since $(1+x) \cdot \frac1{1+x} = 1$, I know that if we applied the operator $(1+D)$ to some function, $\frac{1}{1+D}$ would invert it back to the original function.  However, this still doesn't help me make sense of the meaning of $\frac{1}{1+D}$. For example:   If $1+D$ comes from the differential equation $y'+ay+q$, where does $\frac{1}{1+D}$ come from? In other words, if $(1+D)y = 1\cdot y + Dy = y' + y$, how do we compute $\frac{1}{1+D}y$? (Is it even possible, and how does it relate to integration?) How do we justify the power series of the operator? How do we know there are no convergence issues?
81784
$$\int \frac{1}{1+\tan x}dx,$$  A substitution like $t = \tan x, \;dt = (1+t^2)dx$ etc. immediately comes to mind, but I find this method a bit lengthy with the partial fractions. Is there a more concise solution to this?
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?  $$\int \frac{1}{2+ \sin x + \cos x} ~\mathrm{d}x$$