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+a⁢∫f⁢(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 a≠1a\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 a⁢∫f⁢(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

∫e2⁢x⁢cos⁡(x)⁢𝑑x\int e^{2x}\cos(x)dx

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

∫cos⁡(x)𝑣⁢e2⁢xd⁢u⁢𝑑x=e2⁢x2𝑣⁢cos⁡(x)𝑣−∫e2⁢x2𝑢⁢[−sin⁡(x)]d⁢v⁢𝑑x\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 ∫e2⁢x2⁢[−sin⁡(x)]\int\frac{e^{2x}}{2}[-\sin(x)] by parts.

∫[−sin⁡(x)]𝑢⁢e2⁢x2d⁢v=[−sin⁡(x)]𝑢⁢e2⁢x4𝑣−∫e2⁢x4𝑣⁢[−cos⁡(x)]d⁢u⁢𝑑x\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)⁢e2⁢x⁢𝑑x=e2⁢x2⁢cos⁡(x)−e2⁢x4⁢[−sin⁡(x)]−12⁢∫e2⁢x2⁢cos⁡(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 ∫e2⁢x2⁢cos⁡(x)⁢𝑑x\int\frac{e^{2x}}{2}\cos(x)dx to both sides, giving that

54⁢∫cos⁡(x)⁢e2⁢x⁢𝑑x=e2⁢x2⁢cos⁡(x)+e2⁢x4⁢sin⁡(x)\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 45\frac{4}{5}, we get that

∫cos⁡(x)⁢e2⁢x⁢𝑑x=2⁢e2⁢x⁢cos⁡(x)+e2⁢x⁢sin⁡(x)5\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.