# 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 \sayan integral we know to directly find the value of)

$\int f(x)dx=k_{0}+k_{1}+k_{2}+...+k_{n}+a\int f(x)dx$ (11.12)

It is important that $a\neq 1$ (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

$\int e^{2x}\cos(x)dx$

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

$\int\underset{v}{\cos(x)}\underset{du}{e^{2x}}dx=\underset{v}{\frac{e^{2x}}{2}% }\underset{v}{\cos(x)}-\int\underset{u}{\frac{e^{2x}}{2}}\underset{dv}{[-\sin(% x)]}dx$

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

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

Overall then, we have

$\int\cos(x)e^{2x}dx=\frac{e^{2x}}{2}\cos(x)-\frac{e^{2x}}{4}[-\sin(x)]-\frac{1% }{2}\int\frac{e^{2x}}{2}\cos(x)dx$

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

$\frac{5}{4}\int\cos(x)e^{2x}dx=\frac{e^{2x}}{2}\cos(x)+\frac{e^{2x}}{4}\sin(x)$

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

$\int\cos(x)e^{2x}dx=\frac{2e^{2x}\cos(x)+e^{2x}\sin(x)}{5}$

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