$I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} = ?$

$I = \int\limits_0^\pi {\frac{{2(\pi – x)\sin (\pi – x)}}{{3 + \cos 2(\pi – x)}}dx} $

$ = \int\limits_0^\pi {\frac{{2(\pi – x)\sin x}}{{3 + \cos 2x}}dx} $

$2I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} + \int\limits_0^\pi {\frac{{2(\pi – x)\sin x}}{{3 + \cos 2x}}dx} = \int\limits_0^\pi {\frac{{2\pi \sin x}}{{3 + \cos 2x}}dx} $ Continue reading $I = \int\limits_0^\pi {\frac{{2x\sin x}}{{3 + \cos 2x}}dx} = ?$

$x^2-x-1=0$

$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = ?$

$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = \frac{{({x^2} – 1)({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}}$

$\because {x^2} – x – 1 = 0,{x^2} – 1 = x$

$R(x) = \frac{{x({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^7}(x + 2)}} = \frac{{({x^2} + 1)({x^4} + 1)({x^8} + 1)}}{{{x^6}(x + 2)}}$

$\because {x^2} – x – 1 = 0,{x^2} = x + 1$ Continue reading $x^2-x-1=0$

$R(x)=\frac{{{x^{16}} – 1}}{{{x^8} + 2{x^7}}} = ?$

An $\alpha$-particle (mass 4 amu) and a singly ….

An $\alpha$-particle (mass 4 amu) and a singly charged sulfur ion (mass 32 amu) are initially at rest. They are accelerated through a potential V and then allowed to pass into a region of uniform magnetic field which is normal to the velocities of the particles. Within this region, the $\alpha$-particle and the sulfur ion move in circular orbits of radii $r_{\alpha}$ and $r_S$, respectively. The ratio ($r_S/r_\alpha$) is _____. Continue reading An $\alpha$-particle (mass 4 amu) and a singly ….