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:

1 thought on “Integration by Parts Done Right

  1. Pingback: Eye Of The Tiger | Vroom

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s