Feb 27, 2019 · I cannot find what is the integral of a cumulative distribution function $$\\int G(\\xi)d\\xi$$ I think it should be simple, but I have no idea where else to look for it.

For questions about the properties of integrals. Use in conjunction with (indefinite-integral), (definite-integral), (improper-integrals) or another tag(s) that describe the type of integral being considered. This tag often goes along with the (calculus) tag.

Dec 2, 2014 · What is the integral of $\sin(\cos x)$ ? So glad you asked ! :-) Although the indefinite integral does not possess a closed form, its definite counterpart can be expressed in terms of certain special functions, such as Struve H and Bessel J.

Feb 21, 2018 · I was trying to do this integral $$\int \sqrt{1+x^2}dx$$ I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form of $\int x^m(a+bx^n)^p\,dx$ and I spent a lot of time on it so I would like to see if it can be done this way and where did I go wrong. $$\int(1+x^2 ...

Thus, it cannot be integrated in finite elementary symbols, which is why any answer to the integral requires infinitely many symbols, or a new notation. However, I obviously used some heavy results. If you want to see the proof of these results, I recommend researching differential algebra.

Well, $$\int\frac{2}{x+2} dx= 2\log(x+2)$$ So you have that part. But for: $$\int\frac{1-2x}{x^2 + 1} dx$$ You must further decompose this fraction into partial ...

Sep 27, 2013 · My HW asks me to integrate $\\sin(x)$, $\\cos(x)$, $\\tan(x)$, but when I get to $\\sec(x)$, I'm stuck.

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Oct 7, 2020 · By just checking the left end point only we already see the integral diverges. This is because on the interval $[0,2\pi]$, $$ |\! \sin x| \leq x \implies \left|\frac{1}{\sin x} \right| \geq \frac1x ,$$ and the above leads us to $$ \int_0^{2 \pi} \left| \frac{1}{\sin x} \right| dx \geq \int_0^{2 \pi} \frac1x dx .$$ The later integral diverges, so by comparison test the former also diverges.

$\begingroup$ @user599310, I am going to attempt some pseudo math to show it: $$ I^2 = \int e^-x^2 dx \times \int e^-x^2 dx = Area \times Area = Area^2$$ We can replace one x, with a dummy variable, move the dummy copy into the first integral to get a double integral.