Let $\displaystyle g(x) = \int_0^{x} (x - t)^2 \sin t\,dt \qquad \text{for every real } x.$ The variable $x$ appears twice here: in the upper limit, and inside the integrand. Expanding $(x - t)^2$ and integrating each term by parts works, but it is long, and a sign slip is easy to make somewhere along the way. There is a shorter route, and a much safer one. It never computes $g$ directly until the very end, and it needs only an easy function and the values of $g$ and its first few derivatives at $x = 0$. What is the value of $g(\pi)$?