mathimatikoi.org

\begin{align*} \displaystyle \int_{\pi}^{2\pi}{\frac{\sin x}{e^x-\sin x-\cos x} dx}&=\int_{\pi}^{2\pi}{\frac{1}{2}\Bigl({\frac{e^x-\cos x+\sin x}{e^x-\sin x-\cos x}-1}\Bigr) dx}\\ &=\frac{1}{2}\int_{\pi}^{2\pi}{\frac{e^x-\cos x+\sin x}{e^x-\sin x-\cos x}dx}-\frac{1}{2}\int_{\pi}^{2\pi}{dx}\\...

The \(n\)-term of the sequence \(\{{a_{n}}\}_{n=1}^{\infty}\) with \[a_{n}=\displaystyle\mathop{\prod}_{k=1}^{n}\Bigl({1+\frac{k}{n^2}}\Bigl)\] can also be written as \begin{align*} a_{n}&=\exp(\log(a_{n}))=\exp\Bigl({\log\Bigl[{\textstyle\prod_{k=1}^{n}\bigl({1+\tfrac{k}{n^2}}\bigr)}\Bigr]}\Big...

Find \[\displaystyle\mathop{\lim}\limits_{n\to+\infty}\mathop{\prod}_{k=1}^{n}\Bigl({1+\frac{k}{n^2}}\Bigl)\,.\]

We give a solution which can be found in J. Edwards, A treatise on the integral calculus vol. II : We use that \[\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^2(\tan{x})\,dx}=\frac{\pi^3}{8}\quad(1)\,,\] which is proven here Let \begin{align*} I&=\displaystyle \int_{0}^{\frac{\pi}{2}}{\log^2(\sin...

We supplement the above solution: $$\begin{align*} I(m)&=\int_{0}^{1}{\frac{u^{m-1}}{\sqrt{1-u^2}}\,du}=\frac{\sqrt{\pi}}{2}\frac{\Gamma\bigl({\tfrac{m}{2}}\bigr)}{\Gamma\bigl({\tfrac{m+1}{2}}\bigr)}\quad\Rightarrow\\ \frac{d^2}{dm^2}I(m)&=\int_{0}^{1}{\frac{u^{m-1}\log^2{u}}{\sqrt{1-u^2}}\,...

I guess there are smarter solutions than the one that follows and is based on standard methods of integration: We assume that \(b>a>0\). The case \(b>a>0\) can be treated analogously. The case \(b=a>0\) is easy. By the substitution \(t=\tan{\tfrac{x}{2}}\) we have \(\cos{x}={\displaystyle{\frac{1-t...

Evaluate the integral:

$$\int_{0}^{\frac{\pi}{2}}{\log^2(\sin{x})\,dx}$$

We give calculations for the integrals \(i.\) and \(iii.\) : \(i.\) \begin{align*} \displaystyle \int{\frac{ \tan{x} }{ \sqrt{ a+b\tan^{2}{x} } } \mathrm{d}x}&\mathop{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}\limits^{\begin{subarray}{c} {t\,=\,\tan^2{x}}\\ {\frac{1}{2(1+t)}\,dt\,=\,\tan{x}\,dx}\\ \en...

2nd solution: \begin{align*} \displaystyle\int{\sqrt{\tan^2{x}+2}\;dx}&=\int{\sqrt{\tan^2{x}+1+1}\;dx}\\ &=\int{\sqrt{\frac{1}{\cos^2{x}}+1}\;dx}\\ &=\displaystyle\int{\frac{\sqrt{1+\cos^2{x}}}{\sqrt{\cos^2{x}}}\,dx}\\ &=\int{\frac{\sqrt{2-\sin^2{x}}}{\cos{x}}\,dx}\\ &=\int{\fra...

Evaluate \(\displaystyle\int{\sqrt{\tan^2{x}+2}\;dx}\,.\)