Question

If \(I_{\mathrm{m}, \mathrm{n}}=\int \mathrm{x}^{\mathrm{m}} \cos \mathrm{n} \mathrm{x} \mathrm{dx}(\mathrm{n} \neq 0)\), then show that \(I_{m, n}=\frac{x^{m} \sin n x}{n}+\frac{m x^{m-1} \cos n x}{n^{2}}-\frac{m(m-1)}{n^{2}} I_{m-2, n^{-}}\)

Short answer

Question: Express the integral \(I_{m,n} = \int_0^x t^m \cos nt \, dt\) in terms of \(x^m \sin nx\), \(x^{m-1} \cos nx\), and \(I_{m-2,n}\). Answer: \(I_{m,n} = \frac{x^m \sin nx}{n} + \frac{m x^{m-1} \cos nx}{n^2} - \frac{m(m-1)}{n^2} I_{m-2, n}\)

Step-by-step solution

Integrate by parts

Let \(u = x^m\) and \(v' = \cos{nx}\), so \(u' = m x^{m-1}\) and \(v = \frac{1}{n} \sin{nx}\). We'll apply the integration by parts formula: \(\int u v' \, dx = uv - \int u'v \, dx\). \(I_{m, n} = \int x^m \cos nx \, dx = x^m \cdot\frac{1}{n}\sin nx - \int m x^{m-1}\cdot\frac{1}{n} \sin nx \, dx\)

Simplify the expression

Rewrite the expression to put all constants in front: \(I_{m,n} = \frac{x^m \sin nx}{n} - \frac{m}{n} \int x^{m-1} \sin nx \, dx\)

Integrate by parts again

Now, let's integrate the remaining integral by parts once more. Let \(u = x^{m-1}\) and \(v' = \sin nx\). Then, \(u' = (m-1) x^{m-2}\) and \(v = -\frac{1}{n} \cos nx\). Again, apply the integration by parts formula. \(I_{m,n} = \frac{x^m \sin nx}{n} - \frac{m}{n} \left( -x^{m-1}\cdot\frac{1}{n}\cos nx - \int (m-1) x^{m-2} \cdot\frac{1}{n}\cos nx \, dx\right)\)

Simplify the expression

Simplify the expression and isolate the final integral: \(I_{m,n} = \frac{x^m \sin nx}{n} + \frac{m x^{m-1} \cos nx}{n^2} - \frac{m(m-1)}{n^2} \int x^{m-2} \cos nx \, dx\)

Recognize the integral

Notice that the remaining integral is \(I_{m-2,n}\): \(I_{m,n} = \frac{x^m \sin nx}{n} + \frac{m x^{m-1} \cos nx}{n^2} - \frac{m(m-1)}{n^2} I_{m-2, n}\) This completes the proof.