� Another Reduction Formula: � To compute xn ex dx xn ex dx we derive another reduction formula. We could replace ex by cos x or sin x in this integral and the process would be very similar. Again we’ll use integration by parts to ﬁnd a reduction formula. Here we choose u = xn because u� = nxn−1 is a simpler (lower degree) function. If u = xn then we’ll have to have v � = ex , v = ex . (Note that the antiderivative of v is no more complicated than v � was — another indication that we’ve chosen correctly.) On the other hand, if we used u = ex , then u� = ex would not be any simpler. Performing the integration by parts we get: � � n x n x n−1 x x � ��e � dx = �x��e � − x � �� e � dx. uv � uv If: � Gn (x) = u� v xn ex dx then we get the reduction formula: Gn (x) = xn ex − nGn−1 (x). Let’s illustrate this by computing a few integrals. First we directly compute: � G0 (x) = x0 ex dx = ex + c. Now we can use the reduction formula to conclude that: G1 (x) � So = xex − G0 (x) = xex − ex + c. xex dx = xex − ex + c. Question: How do you know when this method will work? Answer: Good question! The answer is “only through experience and prac tice”. To use this method on an integrand, we need one factor u of the integrand to get simpler when we diﬀerentiate and the other factor v not to get more com plicated when we integrate. We’ve seen how to use integration by parts to derive reduction formulas. We could also ﬁnd these formulas by advanced guessing — guess what the formula should be and then check it. Either method is valid. 1