1^1-3 ھەل قىلىش | مىكروسوفت ماتېماتىكا ھەللىغۇچى

ئاساسىي مەزمۇنغا ئاتلاش

ھەل قىلىش ئەمەلىلەشتۈرۈش قويۇش

تېمىلار

ئالگېبرالىق كىرگۈزگۈچ

ئالگېبرالىق كىرگۈزگۈچ

Trigonometry Inputs

Trigonometry Inputs

Calculus Inputs

Calculus Inputs

Matrix Inputs

ھەل قىلىش ئەمەلىلەشتۈرۈش قويۇش

تېمىلار

ئالگېبرالىق كىرگۈزگۈچ

ئالگېبرالىق كىرگۈزگۈچ

Trigonometry Inputs

Trigonometry Inputs

Calculus Inputs

Calculus Inputs

Matrix Inputs

ھېسابلاش

Tick mark Image

كۆپەيتكۈچى

Tick mark Image

Quiz

تور ئىزدەشتىكى مۇشۇنىڭغا ئوخشاش مەسىلىلەر

https://www.quora.com/How-come-1-2-1-and-1-3-1

https://math.stackexchange.com/questions/2415838/can-one-have-lim-limits-n-to-infty-sum-limits-k-1n10k-n-ak-1-0

https://math.stackexchange.com/questions/1436135/what-is-the-solution-sets-for-5x-2x-5-5x-22x1

https://math.stackexchange.com/q/2890453

From the definition of Cosine Integral : \begin{align} \text{Ci}(x)&=-\int_{x}^{\infty}\frac{\cos{t}}{t}\,dt=\gamma+\log{x}+\int_{0}^{x}\frac{\cos{t}-1}{t}\,dt \\[8mm] I_r&=\int r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr \\[2mm] &\color{blue}{\small\left\{u=\sin\left(\frac{1-3z^2}{r^3}\right)\Rightarrow du=\frac{-3(1-3z^2)}{r^4}\cos\left(\frac{1-3z^2}{r^3}\right) dr,\quad dv=r^2 dr\Rightarrow v=\frac{r^3}{3}\right\}} \\[2mm] I_r&=\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)+(1-3z^2)\int\frac{1}{r}\,\cos\left(\frac{1-3z^2}{r^3}\right)\,dr \\[2mm] &\color{blue}{\small\left\{\frac{d}{dx}\left[\text{Ci}(x)\right]=\frac{\cos{x}}{x}\Rightarrow\frac{d}{dx}\left[\text{Ci}\left(\frac{1-3z^2}{r^3}\right)\right]=\frac{\cos\left(\frac{1-3z^2}{r^3}\right)}{\frac{1-3z^2}{r^3}}\frac{-3(1-3z^2)}{r^4}=-\frac{3}{r}\cos\left(\frac{1-3z^2}{r^3}\right)\right\}} \\[2mm] I_r&=\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)-\frac{1-3z^2}{3}\,\text{Ci}\left(\frac{1-3z^2}{r^3}\right)+const \\[8mm] \color{red}{I}&=\int_{-1}^{+1}\int_{0}^{\infty}r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr\,dz=2\int_{0}^{1}\int_{0}^{\infty}r^2\,\sin\left(\frac{1-3z^2}{r^3}\right)\,dr\,dz\quad\color{blue}{\left\{\rightarrow z^2\right\}} \\[2mm] &=2\int_{0}^{1}\left[\frac{r^3}{3}\,\sin\left(\frac{1-3z^2}{r^3}\right)-\frac{1-3z^2}{3}\,\text{Ci}\left(\frac{1-3z^2}{r^3}\right)\right]_{r=0}^{r=\infty}\,dz \\[2mm] &=2\int_{0}^{1}\left[\frac{1-3z^2}{3}\left(1-\gamma-\log\left(1-3z^2\right)\right)\right]\,dz \quad\color{blue}{\left\{\rightarrow\text{Ci}(x)=\gamma+\log{x}-\text{Cin}(x)\right\}} \\[2mm] &=\frac{2}{3}(1-\gamma)\int_{0}^{1}\left(1-3z^2\right)\,dz\,-\,\frac{2}{3}\int_{0}^{1}\left(1-3z^2\right)\left(\log\left(1-3z^3\right)\right)\,dz\qquad\color{blue}{\left\{\rightarrow\text{IBP}\right\}} \\[2mm] &=\frac{2}{3}(1-\gamma)\left[z-z^3\right]_{0}^{1}\,+\,\frac{2}{3}\left[\frac{2}{3}z+\left(z-z^3\right)\left(\frac{2}{3}+\log\left(1-3z^2\right)\right)-\frac{4}{3\sqrt{3}}\text{arctanh}(\sqrt3z)\right]_{0}^{1} \\[2mm] &=\color{red}{\frac{4}{9}-\frac{8}{9\sqrt{3}}\,\text{arctanh}{\sqrt{3}}}\,\approx{0.106513} \end{align}

تەڭ بەھرىمان بولۇش

قىسقۇچقا كۆچۈرۈلگەن

1^{-2}

1 دىن 3 نى ئېلىپ -2 نى چىقىرىڭ.

1

1 نىڭ -2-دەرىجىسىنى ھېسابلاپ 1 نى چىقىرىڭ.

مىساللار

تۆت تەرەپ تەڭلىمىسى

سىزىقلىق تەڭلىمە

پەرقلەندۈرۈش

بىرىكتۈرۈش