MA005: Calculus I

A recursive formula in the table is one that gives one antiderivative in terms of another antiderivative. Usually the new antiderivative is simpler than the original one. For example, table entry #19 gives the pattern for the antiderivative of $\sin ^{\mathrm{n}}(\operatorname{ax})$ in terms of the antiderivative of $\sin ^{n-2}(a x)$. If we start with the integral of $\sin ^{5}(x)$, then we can use the table to get an answer in terms of the integral of $\sin ^{3}(x)$. Using the recursion formula again, we can get the integral of $\sin ^{3}(x)$ in terms of the integral of $\sin (x)$, which is easy to integrate.

Table entry 19: $\int \sin ^{n}(a x) d x=\frac{-\sin ^{n-1}(a x) \cdot \cos (a x)}{n a}+\frac{n-1}{n} \int \sin ^{n-2}(a x) d x$.

Example 5: Use the given recursive formula to evaluate $\int \sin ^{3}(5 x) \mathrm{dx}$.

Solution: In this example $\mathrm{n}=3$ and $\mathrm{a}=5$. Then

$\begin{aligned} \int \sin ^{3}(5 x) \mathrm{dx} &=\frac{-\sin ^{2}(5 x) \cdot \cos (5 x)}{3 \cdot 5}+\frac{2}{3} \int \sin (5 x) \mathrm{dx} \\ &=\frac{-\sin ^{2}(5 x) \cos (5 x)}{15}-\frac{2}{3} \frac{1}{5} \cos (5 x)+\mathrm{C} \end{aligned}$

Practice 5: Use the recursion formula in the table to evaluate $\int \cos ^{3}(7 x) \mathrm{d} x$.