Evaluating $\\int_1^{\\sqrt{2}} \\frac{\\arctan(\\sqrt{2-x^2})}{1+x^2 ...
3 days ago · I was recently trying to compute the value of the integral $$\int_1^ {\sqrt {2}} \frac {\arctan (\sqrt {2-x^2})} {1+x^2}\,\mathrm dx.$$ I’ve tried differentiation under the integral sign, …
contest math - Evaluating $4^ {5 \log_ {4\sqrt {2}} (3-\sqrt {6}) - 6 ...
May 9, 2025 · Here's another, seemingly monstrous question from a JEE Advanced preparation book. Evaluate the following expression: $$4^ {5 \log_ {4\sqrt {2}} (3-\sqrt {6}) - 6\log ...
algebra precalculus - Evaluating $\frac {1} {a^ {2025}}+\frac {1} {b ...
Feb 21, 2025 · Well, the image equation is a different equation? One has $\frac1 {2024}$ on the right, and the other has $2024$ on the right?
Evaluating $\int_ {0}^ {1} \int_ {0}^ {1} \int_ {0}^ {1} \sqrt { {x^2+y ...
Aug 27, 2019 · I would like to know how to evaluate the following triple integral with the help of spherical coordinates $$\int_ {0}^ {1} \int_ {0}^ {1} \int_ {0}^ {1} \sqrt { {x^2+y^2+z^2}} \,dx \,dy\, …
calculus - Evaluating $\int {\frac {x^ {14}+x^ {11}+x^5}
Jul 2, 2025 · The following question is taken from JEE practice set. Evaluate $\displaystyle\int {\frac {x^ {14}+x^ {11}+x^5} {\left (x^6+x^3+1\right)^3}} \, \mathrm dx$. My ...
Evaluating $ \\lim_{x \\to 0} \\frac{e - (1 + 2x)^{1/2x}}{x} $ without ...
Sep 11, 2024 · The following is a question from the Joint Entrance Examination (Main) from the 09 April 2024 evening shift: $$ \lim_ {x \to 0} \frac {e - (1 + 2x)^ {1/2x}} {x} $$ is equal to: (A) …
integration - Evaluating $\iiint z (x^2+y^2+z^2)
Jul 29, 2020 · Spherical Coordinate Homework Question Evaluate the triple integral of $f (x,y,z)=z (x^2+y^2+z^2)^ {−3/2}$ over the part of the ball $x^2+y^2+z^2\le 81$ defined by ...
algebra precalculus - Evaluating $\csc\left (\frac {\pi} {18}\right ...
Nov 27, 2023 · Noting that $$ \frac {1} {2}=\sin \left (3\left (\frac {\pi} {18}\right)\right)=3 \sin \left (\frac {\pi} {18}\right)-4 \sin ^3\left (\frac {\pi} {18}\right ...
Evaluating $\sum_ {i=1}^ {\infty}\frac { (i\ln 2)^i} {2^ii!}$
Dec 26, 2024 · I seek the proof of the evaluation to the sum $$\sum_ {i=1}^ {\infty}\frac { (i\ln 2)^i} {2^ii!} = \frac {1} {1-\ln2}-1 \approx 2.25889.$$ It is almost a power series ...
Evaluating $\\lim_{n\\to\\infty}\\left( \\frac{\\cos\\frac{\\pi}{2n ...
Jan 24, 2025 · Since the OP solve his/her problem, I just as well complete the solution: \begin {align} \frac {1} {n+1}\sum^n_ {k=1}\cos\left (\tfrac {k\pi} {2n}\right)&=\frac {n ...