Question

In Exercises \(47-50,\) use integration by parts to establish the reduction formula. $$\int x^{n} \sin x d x=-x^{n} \cos x+n \int x^{n-1} \cos x d x$$

Short answer

The reduction formula \(\int x^{n} \sin x dx = -x^{n} \cos x + n \int x^{n-1} \cos x dx\) has been successfully established using integration by parts.

Step-by-step solution

Identify u and dv

In the formula \(\int u dv = u*v - \int v du\), we need to specify \(u\) and \(dv\) first. Here \(u = x^{n}\) and \(dv = \sin x dx\). The choice is made this way because the derivative of \(x^{n}\) (which will be required for du) simplifies as we decrease the power of x, and the integral of \(\sin x\) can be easily found.

Compute dv and du

Now, find the derivative of \(u\). \(du = n x^{n-1} dx\). Then find the antiderivative of \(dv\) to get \(v\). \(v = \int dv = -\cos x\).

Apply the integration by parts formula

Substitute \(u\), \(v\), and \(du\) into the formula for integration by parts. \(\int x^{n}\sin x dx = u*v - \int v du = -x^{n} \cos x + \int n x^{n-1} \cos x dx\). Note that the sign within the definite integral changes because of the minus outside.