November 2020

Integration by parts is a technique for evaluating indefinite integral whose integrand is a product of two functions. It is based on the fact that

If two functions $u(x), v(x)$ are differentiable on an interval $I$ and $\int u(x)v'(x)\;dx$ exists, then

$\int u'(x) v(x)\;dx = u(x)\cdot v(x)-\int u(x)\cdot v'(x)\;dx\quad\quad\quad(1)$

It is not difficult to see that (1) is true:

Provide $\int u(x)\cdot v(x)'\;dx$ exist, let

$u = u(x), v = v(x)$

and

$R = u\cdot v - \int u\cdot v'\; dx$ .

By the rules of differentiation (see “Some rules of differentiation”),

$R' = (u\cdot v)' - (\int u \cdot v'\;dx)'= u'\cdot v +u\cdot v' - u\cdot v' = u'\cdot v$ .

i.e.,

$R = \int u'\cdot v\;dx$

or,

$\int u'\cdot v\;dx= u\cdot v - \int u\cdot v'\;dx$

The key to apply this technique successfully is to choose proper $u, v$ so that the integrand in the original integral can be expressed as $u'\cdot v$ and, $\int u\cdot v'\;dx$ is easier to evaluate than $\int u'\cdot v\;dx$ .

To evaluate $\int \log(x)\;dx$ , we let $u = x, v=\log(x)$ to get

$\int \log(x)\;dx = \int 1\cdot \log(x)\;dx$

$= \int (x)' \cdot \log(x) \;dx$

$= x\cdot\log(x) - \int x\cdot (\log(x))'\;dx$

$= x\log(x) - \int x \cdot \frac{1}{x}\;dx$

$= x\log(x) - \int 1\;dx$

$= x\log(x) -x$

In fact, for $n \ge 0$ ,

$\int x^n \log(x)\;dx = \int (\frac{x^{n+1}}{n+1})'\log(x)\;dx$

$= \frac{x^{n+1}}{n+1}\log(x)-\int \frac{x^{n+1}}{n+1}(\log(x))'\;dx$

$= \frac{x^{n+1}}{n+1}\log(x)-\int\frac{x^{n+1}}{n+1}\cdot \frac{1}{x}\;dx$

$= \frac{x^{n+1}}{n+1}\log(x)-\frac{1}{n+1}\int x^n\;dx$

$= \frac{x^{n+1}}{n+1}\log(x)-\frac{x^{n+1}}{(n+1)^2}$ .

The following example of integrating a rather ordinary-looking expression offers unexpected difficulties and surprises:

$\int \sqrt{1-x^2}\;dx= \int x' \sqrt{1-x^2}\;dx$

$= x\sqrt{1-x^2}-\int x(\sqrt{1-x^2})'\;dx$

$= x\sqrt{1-x^2}- \int x\cdot \frac{1}{2}\cdot\frac{-2x}{\sqrt{1-x^2}}\;dx$

$= x\sqrt{1-x^2} +\int \frac{x^2}{\sqrt{1-x^2}}\;dx$

$= x\sqrt{1-x^2} + \int \frac{1-1+x^2}{\sqrt{1-x^2}}\;dx$

$= x\sqrt{1-x^2} + \int \frac{1}{\sqrt{1-x^2}}\;dx -\int \frac{1-x^2}{\sqrt{1-x^2}}\;dx$

$\overset{\frac{1-x^2}{\sqrt{1-x^2}}=\sqrt{1-x^2}}{= }x\sqrt{1-x^2}+\arcsin(x) - \boxed{\int\sqrt{1-x^2}\;dx}$

Notice the original integral (boxed) appears on the right of the “=” sign. However, this is not an indication that we have reached a dead end. To the contrary, after it is combined with the left side, we have

$2\int\sqrt{1-x^2}\;dx = x\sqrt{1-x^2} + \arcsin(x)$

and so,

$\int \sqrt{1-x^2}\;dx = \frac{1}{2}(x\sqrt{1-x^2} + \arcsin(x))$

Sometimes, successive integration by parts is required to complete the integration. For example,

$\int e^x \cos(x)\;dx = \int (e^x)' \cos(x)\;dx$

$= e^x \cos(x)-\int e^x (\cos(x)' \; dx$

$= e^x \cos(x) +\int e^x \sin(x)\; dx$

$= e^x\cos(x)+\int (e^x)' \sin(x)\;dx$

$= e^x \cos(x) + (e^x \sin(x)-\int e^x (\sin(x))' \;dx)$

$=e^x \cos(x) + e^x \sin(x) - \boxed{\int e^x \cos(x)\;dx}$ .

Hence,

$2\int e^x\cos(x)\;dx = e^x\cos(x) + e^x\sin(x)$

i.e.,

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

Exercise-1 Evaluate

1) $\int \log^2(x)\;dx$

2) $\int x\log(\frac{1-x}{1+x})\;dx$

3) $\int \log(x+\sqrt{1+x^2})\; dx$

While Maxima choked:

Mathematica came through:

4) $\int \frac{x}{\sqrt{1+2x}}\;dx$

5) $\int x\cdot \arctan(x)\;dx$

6) $\int \frac{x^2}{(1+x^2)^2}\;dx$

7) $\int \frac{x^2\cdot e^x}{(x+2)^2}\;dx$

Both Maxima and Mathematica came though:

Shown in Fig. 1 is a pile of $n$ cylinders that approximates one half of a sphere.

Fig. 1

From Fig. 1, we see $r_i^2 = r^2-h_i^2$ .

If $\Delta V_i$ denotes the volume of $i^{th}$ cylinder in the pile, then

$\Delta V_i = \pi \cdot r_i^2 \cdot \frac{r}{n}$

$= \pi (r^2-h_i^2) \cdot \frac{r}{n}$

$= \pi (r^2-(i\cdot\frac{r}{n})^2)\cdot \frac{r}{n}$

$= \frac{\pi r^3}{n} (1-(\frac{i}{n})^2)$ .

Let

$V_{*}=\sum\limits_{i=1}^{n}\Delta V_i$ ,

we have

$V_{*}= \sum\limits_{i=1}^{n}\frac{\pi r^3}{n} (1-(\frac{i}{n})^2)$

$= \frac{\pi r^3}{n}\sum\limits_{i=1}^{n}(1-(\frac{i}{n})^2)$

$= \frac{\pi r^3}{n}(\sum\limits_{i=1}^{n}1-\sum\limits_{i=1}^{n}(\frac{i}{n})^2)$

$= \frac{\pi r^3}{n}(n-\frac{1}{n^2}\sum\limits_{i=1}^{n}i^2)$

$= \frac{\pi r^3}{n}(n-\frac{n(n+1)(2n+1)}{6n^2})\quad\quad\quad($ see “Little Bird and a Recursive Generator“ $)$

$= \pi r^3(1-\frac{(n+1)(2n+1)}{6n^2})$

$= \pi r^3(1-\frac{2n^2+3n+1}{6n^2})$

$= \pi r^3(1-\frac{2+\frac{3}{n}+\frac{1}{n^2}}{6})$ .

As $n \rightarrow \infty,$

$V_* \rightarrow \pi r^3 (1-\frac{1}{3}) = \frac{2}{3}\pi r^3=V_{\frac{1}{2}sphere}.$

Therefore, $V_{sphere}= 2\cdot V_{\frac{1}{2}sphere}$ gives

$V_{sphere} = \frac{4}{3}\pi r^3\quad\quad\quad(1-1)$

Similarly, we approximate a cone

by a stack of $n$ cylinders:

Fig. 2

Since

$h_i = i\cdot\frac{h}{n}\quad\quad\quad(2-1)$

and

$\frac{h_i}{h} = \frac{r_i}{r}$ ,

we have

$r_i = \frac{r\cdot h_i}{h}\overset{(2-1)}{=}\frac{r}{h}\cdot i\frac{h}{n}=\frac{r\cdot i}{n}.\quad\quad\quad(2-2)$

If $\Delta V_i$ denotes the $i^{th}$ cylinder, then

$\Delta V_i = \pi\cdot r_i^2\cdot\frac{h}{n}\overset{(2-2)}{=}\frac{\pi r^2 i^2 h}{n^3}$ .

Let

$V_* = \sum\limits_{i=1}^{n}\Delta V_i,$

we have

$V_*= \frac{\pi r^2 h}{n^3}\sum\limits_{i=1}^{n} i^2$

$= \frac{\pi r h}{n^3} \cdot \frac{2n^3+3n^2+n}{6}$

$= \pi r^2 h (\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2})$ .

Since as $n \rightarrow \infty$ , $V_*\rightarrow \frac{1}{3}\pi r^2 h$ , we have

$V_{cone} = \frac{1}{3}\pi r^2 h.\quad\quad\quad(2-3)$

We now derive the formula for a sphere’s surface area.

Let us approximate a sphere by many cones:

Fig. 3

From Fig. 3, we see that as $\Delta A_i \rightarrow 0$ ,

$h \rightarrow r, \;\sum\limits_{i}\Delta A_i \rightarrow A_{sphere}$ .

Consequently,

$\sum\limits_{i} \Delta V_i \overset{(2-3)}{=} \sum\limits_{i} \frac{1}{3}\Delta A_i h = \frac{1}{3}h \sum\limits_{i}\Delta A_i \rightarrow \frac{1}{3}rA_{sphere}$