# A calculus of the absurd

#### 11.4 Integral arithmetic

This technique goes by different names, but integral arithmetic captures the basic idea pretty well; sometimes it is very helpful to treat integrals as algebraic objects in order to find their value.

A very common example of this is where, by integrating $$f(x)$$ (or any other integrable function) with respect to $$x$$, we can arrive with an equation of the form (here we define $$k$$ to stand for “an integral we know to directly find the value of”)

\begin{equation} \int f(x) dx = k_0 + k_1 + k_2 + ... + k_n + a \int f(x) dx \end{equation}

It is important that $$a \ne 1$$ (because if $$a$$ is equal to one then we cannot solve for $$\int f(x) dx$$), in which case we can just subtract $$a \int f(x) dx$$ from both sides, to solve for $$\int f(x) dx$$.

• Example 11.4.1 Find the value of

\begin{equation*} \int e^{2x} \cos (x) dx \end{equation*}

Solution: Start by integrating by parts (as in Section 11.2)

\begin{equation*} \int \underset {v}{\cos (x)} \underset {du}{e^{2x}} dx = \underset {v}{\frac {e^{2x}}{2}}\underset {v}{\cos (x)} - \int \underset {u}{\frac {e^{2x}}{2}} \underset {dv}{[-\sin (x)]} dx \end{equation*}

Then integrate $$\int \frac {e^{2x}}{2}[-\sin (x)]$$ by parts.

\begin{equation*} \int \underset {u}{[-\sin (x)]}\underset {dv}{\frac {e^{2x}}{2}} = \underset {u}{[-\sin (x)]}\underset {v}{\frac {e^{2x}}{4}} - \int \underset {v}{\frac {e^{2x}}{4}} \underset {du}{[-\cos (x)]} dx \end{equation*}

Overall then, we have

\begin{equation*} \int \cos (x) e^{2x} dx = \frac {e^{2x}}{2}\cos (x) - \frac {e^{2x}}{4} [-\sin (x)] - \frac {1}{2} \int \frac {e^{2x}}{2} \cos (x) dx \end{equation*}

And we can add $$\int \frac {e^{2x}}{2} \cos (x) dx$$ to both sides, giving that

\begin{equation*} \frac {5}{4} \int \cos (x) e^{2x} dx = \frac {e^{2x}}{2}\cos (x) + \frac {e^{2x}}{4}\sin (x) \end{equation*}

and then after multiplying both sides by $$\frac {4}{5}$$, we get that

\begin{equation*} \int \cos (x) e^{2x} dx = \frac {2e^{2x}\cos (x) + e^{2x}\sin (x)}{5} \end{equation*}

Integrating by parts can get really messy - good presentation is key.