 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2016/discrete-mathematics-continuous-mathematics-and-probability/5/a/ii/solution

u = x

and

d v = sin (μx) d x

. Part (i) gives

v = - cos (μx) / μ

, so

\int x sin (μx) d x = - \frac{x c o s ( μx )}{μ} + \frac{1}{μ} \int cos (μx) d x = - \frac{x cos ( μx )}{μ} + \frac{sin ( μx )}{μ ^{2}} + C .

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