How would one evaluate \[\int_0^{\frac{2\pi n}{k}}\,\frac{a \sin{(kx)}}{(a^2 + x^2)^{3/2}}\,dx\,?\]
Siddhant has suggested a very interesting approach that I will offer a slight variant of. First, introduce two angle variables, the constant \(\alpha = ka\) and the variable \(\theta = kx\). In terms of these new variables, the integral becomes
As first suggested by Nikhil Tilak, begin by integrating by parts, letting \(u = \sin{(kx)}\) and \(dv = (a^2 + x^2)^{-3/2} dx\). Then \(du = k\cos{(kx)} dx\), and \(v\) can be found by making the change of variables to \(x = a\tan{\theta}\), \(dx = a\sec^2{(\theta)} d\theta\), yielding eventually \(v = \frac{x}{a^2\sqrt{a^2+x^2}}\). Denoting the integral by \(I\), it thereby reduces to
\[I = \frac{ x \sin{(kx)} }{ a(a^2+x^2)^{1/2} }\Big\vert_{0}^{\frac{2\pi n}{k}} - \frac{k}{a}\,\int_{0}^{\frac{2\pi n}{k}}\,\frac{x \cos{(kx)} }{ (a^2 + x^2)^{1/2} }\,dx,\]
\[I = - \frac{k}{a}\,\int_{0}^{\frac{2\pi n}{k}}\,\frac{ x \cos{(kx)} }{ (a^2 + x^2)^{1/2} }\,dx,\]
since the integrated terms vanish at the end points.
Siddhant has suggested a very interesting approach that I will offer a slight variant of. First, introduce two angle variables, the constant \(\alpha = ka\) and the variable \(\theta = kx\). In terms of these new variables, the integral becomes
As first suggested by Nikhil Tilak, begin by integrating by parts, letting \(u = \sin{(kx)}\) and \(dv = (a^2 + x^2)^{-3/2} dx\). Then \(du = k\cos{(kx)} dx\), and \(v\) can be found by making the change of variables to \(x = a\tan{\theta}\), \(dx = a\sec^2{(\theta)} d\theta\), yielding eventually \(v = \frac{x}{a^2\sqrt{a^2+x^2}}\). Denoting the integral by \(I\), it thereby reduces to
\[I = \frac{ x \sin{(kx)} }{ a(a^2+x^2)^{1/2} }\Big\vert_{0}^{\frac{2\pi n}{k}} - \frac{k}{a}\,\int_{0}^{\frac{2\pi n}{k}}\,\frac{x \cos{(kx)} }{ (a^2 + x^2)^{1/2} }\,dx,\]
\[I = - \frac{k}{a}\,\int_{0}^{\frac{2\pi n}{k}}\,\frac{ x \cos{(kx)} }{ (a^2 + x^2)^{1/2} }\,dx,\]
since the integrated terms vanish at the end points.