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)f(x) (or any other integrable function) with respect to xx, we can arrive with an equation of the form (here we define kk to stand for \sayan integral we know to directly find the value of)

f(x)𝑑x=k0+k1+k2++kn+af(x)𝑑x\int f(x)dx=k_{0}+k_{1}+k_{2}+...+k_{n}+a\int f(x)dx (11.12)

It is important that a1a\neq 1 (because if aa is equal to one then we cannot solve for f(x)𝑑x\int f(x)dx), in which case we can just subtract af(x)𝑑xa\int f(x)dx from both sides, to solve for f(x)𝑑x\int f(x)dx.

Example 11.4.1

Find the value of

e2xcos(x)𝑑x\int e^{2x}\cos(x)dx

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

cos(x)𝑣e2xdu𝑑x=e2x2𝑣cos(x)𝑣e2x2𝑢[sin(x)]dv𝑑x\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 e2x2[sin(x)]\int\frac{e^{2x}}{2}[-\sin(x)] by parts.

[sin(x)]𝑢e2x2dv=[sin(x)]𝑢e2x4𝑣e2x4𝑣[cos(x)]du𝑑x\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

cos(x)e2x𝑑x=e2x2cos(x)e2x4[sin(x)]12e2x2cos(x)𝑑x\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 e2x2cos(x)𝑑x\int\frac{e^{2x}}{2}\cos(x)dx to both sides, giving that


and then after multiplying both sides by 45\frac{4}{5}, we get that


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