Monthly Archives: November 2020

Integration by Parts Done Right

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:

In the spirit of Archimedes’ method of exhaustion

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}

and,

\sum\limits_{i} \Delta V_i \rightarrow V_{sphere} \overset{(1-1)}{=} \frac{4}{3}\pi r^3.

It follows that

\frac{1}{3}r A_{sphere} = \frac{4}{3}\pi r^3.

Hence,

A_{sphere} = 4 \pi r^2.