mathimatikoi.org

\(I=\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-4\,\sin{x}\,\cos{x}+2\,\sin^2{x}}\,dx}.\) Substituting \(t=\tan{x}\), we have \(\sin{x}=\displaystyle\frac{t}{\sqrt{1+t^2}}\), \(\cos{x}=\displaystyle\frac...

Prove that
$$\displaystyle\int{\frac{\cos{x}+\sin{x}}{5\,\cos^2{x}-2\,\sin({2x})+2\,\sin^2{x}}\,dx}=\frac{3}{5}\,\arctan({\sin{x}-2\,\cos{x}})+\frac{\sqrt{6}}{60}\,\ln\Bigl|{\tfrac{\sqrt{6}+2\,\sin{x}+\cos{x}}{\sqrt{6}-2\,\sin{x}-\cos{x}}}\Bigr|$$

... and an elementary way: \begin{align*} & \displaystyle\mathop{\sum}\limits_{n=1}^{\infty}{n^{a}\bigl({\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\,}\bigr)}=\\ & \displaystyle\sqrt{2}-2+\mathop{\sum}\limits_{n=2}^{\infty}{n^{a}\biggl({\frac{1}{\sqrt{n+1}+\sqrt{n}}-\frac{1}{\sqrt{n}+\sqrt{n-1}}}\biggr)...

Examine if the series \[\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}{(-1)^{n+1}\cos\tfrac{1}{n}}\] converges.

Examine the convergence of the series \[\displaystyle\mathop{\sum}\limits_{n=1}^{\infty}{n^{a}\bigl({\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1}\,}\bigr)}\] for all real numbers \(a\) .

a) Prove that $$\displaystyle\frac{1}{2}\cdot\frac{3}{4}\cdot\ldots\cdot\frac{99}{100}<\frac{1}{10}\,.$$
b) Examine if, for every \(n\in\mathbb{N}\,,\) holds $$\displaystyle\mathop{\prod}\limits_{k=1}^{n}\frac{2k-1}{2k}<\frac{1}{\sqrt{2n}}\,.$$

For the bijective and continuous function \( h\,: \mathbb{R}\longrightarrow\mathbb{R}\) holds: $$h(x)+h^{ - 1}(x) = 2x \quad (1)\,, \;{\text{for all}} \; x\in\mathbb{R}\,.$$ Because \(h\) is bijective and continuous is strictly monotonic and the inverse function \(h^{-1}\) has the same monotony. But...

c) For the sequence \(\{{a_{n}}\}_{n=1}^{\infty}\) of real numbers holds $$0<a_{1}<a_2\quad {\text{ and}}\quad a_{n+1}=\sqrt{a_{n}\,a_{n-1}}\, , \quad(1) \quad n\geqslant2\, .$$ From \((1)\) we have \begin{alignat*}{2} \log a_{n+1}& =\log\sqrt{a_{n}\,a_{n-1}}=\log\bigl({\sqrt{a_{n}}\,\sqrt{a_{n-...

We use that \[\displaystyle\mathop{\sum}\limits_{k=1}^n\varphi(k)=\frac{3n^2}{\pi^2}+{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)\] Because \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}\frac{1}{n^2}\,{\cal{O}}\big(n\log^{2/3}n\,\log^{4/3}(\log{n})\big)=0\quad (*)\] we have that \begin{al...

\begin{align*} \int_0^1{\frac{\log(1-x)\log{x}}{1-x}\,dx}&\mathop{=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,1-x}\\ {-dt\,=\,dx} \\ \end{subarray}} \,\int_0^1{\frac{\log{t}\,\log(1-t)}{t}\,dt}\\ &=-\int_0^1{\frac{\log{t}}{t}\mathop{\sum}\limits_{n=1}^{\infty}\frac{t^n}{n}\,dt}\\ &a...