Question

In Exercises \(57-62,\) use integration by parts to verify the formula. (For Exercises \(57-60,\) assume that \(n\) is a positive integer. \()\) $$ \int x^{n} \sin x d x=-x^{n} \cos x+n \int x^{n-1} \cos x d x $$

Short answer

The proof shows that the original formula \(\int x^n sinx dx = -x^n cosx + n \int x^{n-1} cosx dx\) is correct.

Step-by-step solution

Define the Function Parts

Integration by parts follows the formula \(\int u dv=uv−\int v du\), where \(u\) is one part of the function and \(dv\) is the other part. In our integration, \(x^n\) is chosen as \(u\), and \(sinx dx\) as \(dv\). So differentiate \(u\) to get \(du = n*x^{n-1} dx\) and integrate \(dv\) to get \(v =-cosx\).

Apply Integration by Parts Formula

Now plug \(u\), \(du\), \(v\), and \(dv\) into the integration by parts formula \(\int u dv=uv−\int v du\). This leads to: \(\int x^n sinx dx = -x^n* cosx - \int -cosx * n*x^{n-1} dx\), which simplifies to: \(-x^n cosx + n \int x^{n-1} cosx dx\).

Compare Result with Original Integration

Compare the resulting formula with the original formula given in the question: \(-x^n cosx + n \int x^{n-1} cosx dx = -x^n cosx + n \int x^{n-1} cosx dx\). The two formulas match, therefore the verification is done. We just demonstrated that: \(\int x^n sinx dx = -x^n cosx + n \int x^{n-1} cosx dx\).