Wikipedia:Reference desk/Archives/Mathematics/2021 February 13

= February 13 =

Integration
How to integrate sin 4 x.cos ^ { 3 } x d x — Preceding unsigned comment added by 2402:4000:11CA:BC1F:2:2:1D15:FEA2 (talk) 13:14, 13 February 2021 (UTC)
 * First use the power reduction formula for $$\cos^3x$$ and then the product to sum formula for $$\sin x_1\cos x_2$$. -Abdul Muhsy (talk) 14:02, 13 February 2021 (UTC)


 * Observe that $$d\cos x = -\sin x\,dx.$$ This can be exploited in this specific case for a change of variable, giving a slightly less laborious calculation. Use the double-angle formulas $$\sin 2x=2\sin x\cos x$$ and $$\cos 2x=2\cos^2x-1$$ repeatedly to rewrite $$\sin 4x\cdot\cos^3x$$ as $$4\sin x\cos x(2\cos^2x-1)\cdot\cos^3x =$$ $$\sin x(8\cos^6x-4\cos^4x).$$ Then
 * $$~\int \sin 4x\cdot\cos^3x\,dx$$
 * $$=\int \sin x(8\cos^6x-4\cos^4x)\,dx$$
 * $$=\int -(8\cos^6x-4\cos^4x)\cdot -\sin x\,dx,$$
 * $$=\int -(8\cos^6x-4\cos^4x)\,d\cos x,$$
 * $$=\int -(8c^6-4c^4)\,dc,$$ where $$c=\cos x, $$
 * $$=-(\frac{8}{7}c^7-\frac{4}{5}c^5)$$
 * $$=-(\frac{8}{7}\cos^7x-\frac{4}{5}\cos^5x).$$
 * --Lambiam 16:33, 13 February 2021 (UTC)
 * Just for fun, another method (expanding on User:Abdul Muhsy's suggestion):
 * $$\int \sin 4x\cdot\cos^3x\,dx$$
 * $$=\int \sin 4x\cdot\cos x\cdot\cos x\cdot\cos x\,dx$$
 * $$=\frac{1}{2}\int (\sin 3x+\sin 5x)\cdot\cos x\cdot\cos x\,dx$$
 * $$=\frac{1}{4}\int (\sin 2x+2\sin 4x+\sin 6x)\cdot\cos x\,dx$$
 * $$=\frac{1}{8}\int (\sin x+3\sin 3x+3\sin 5x+\sin 7x)\,dx$$
 * $$=-\frac{1}{8}(\cos x+\frac{3\cos 3x}{3}+\frac{3\cos 5x}{5}+\frac{\cos 7x}{7})$$
 * --RDBury (talk) 13:03, 14 February 2021 (UTC)


 * Brute force:


 * $$\begin{align}\int \sin(4 x) \cos^{3}(x) dx & = \int \frac{e^{4 i x} - e^{- 4 i x}}{2 i} \left( \frac{e^{i x} + e^{- i x}}{2}\right)^{3} dx \\

& = \int \frac{1}{16 i} (e^{7 i x} + 3 e^{5 i x} + 3 e^{3 i x} + e^{i x} - e^{- i x} - 3 e^{- 3 i x} - 3 e^{- 5 i x} - e^{- 7 i x}) dx \\ & = - \frac{1}{16} \left( \frac{e^{7 i x}}{7} + 3 \frac{e^{5 i x}}{5} + e^{3 i x} + e^{i x} + e^{- i x} + e^{- 3 i x} + 3 \frac{e^{- 5 i x}}{5} + \frac{e^{- 7 i x}}{7}\right) + C \\ & = - \frac{1}{8} \left( \frac{\cos(7 x)}{7} + \frac{3 \cos(5 x)}{5} + \cos(3 x) + \cos(x)\right) + C \end{align}$$


 * (though $$- \frac{8}{7} \cos^{7}(x) + \frac{4}{5} \cos^{5}(x)$$ is surely a neater way of expressing this). catslash (talk) 19:18, 14 February 2021 (UTC)


 * I can't make catslash's multiple angles form match the powers form. —Tamfang (talk) 02:22, 15 February 2021 (UTC)
 * The calculations above constitute a proof of their equality. --Lambiam 12:09, 15 February 2021 (UTC)


 * They could differ by a constant (they do not)
 * $$\scriptstyle{\cos(7 x) = 64 \cos^{7}(x) - 112 \cos^{5}(x) + 56 \cos^{3}(x) - 7 \cos(x)}$$
 * $$\scriptstyle{\cos(5 x) = 16 \cos^{5}(x) - 20 \cos^{3}(x) + 5 \cos(x)}$$
 * $$\scriptstyle{\cos(3 x) = 4 \cos^{3}(x) - 3 \cos(x)}$$


 * $$\scriptstyle{\cos^{7}(x) = \frac{1}{64} (\cos(7 x) + 7 \cos(5 x) + 21 \cos(3 x) + 35 \cos(x))}$$
 * $$\scriptstyle{\cos^{5}(x) = \frac{1}{16} (\cos(5 x) + 5 \cos(3 x) + 10 \cos(x))}$$
 * $$\scriptstyle{\cos^{3}(x) = \frac{1}{4} (\cos(3 x) + 3 \cos(x))}$$
 * catslash (talk) 13:28, 15 February 2021 (UTC)
 * That they don't differ by a constant can be made obvious by setting $x=\pi/2.$ --Lambiam 14:54, 15 February 2021 (UTC)


 * I made a blunder in expanding the multiple angle form. I'll try again. —Tamfang (talk) 01:37, 18 February 2021 (UTC)