Integral Calculus : Integration by Parts

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Integration by Parts

what you'll learn...

Overview

Integration by Parts :

» $\int f (x) g (x) d x$ $int f(x) g(x) dx$ $= f (x) \int g (x) d x$ $= f(x) intg(x)dx$ $- \int [\int g (x) d x] [\frac{d}{d x} f (x)] d x$ $- int [int g(x) dx] [d/(dx)f(x)] dx$

→ $\int e^{x} (h (x) + h' (x)) d x$ $int e^x (h(x)+h′(x)) dx$ $= e^{x} h (x) + c$ $=e^x h(x) + c$

part by part

Integration of functions that are product of sub-functions is hard to find result for. For example, $f (x) = x^{2}$ $f(x)=x^2$ and $g (x) = \sin x$ $g(x)=sinx$ can be individually integrated. But integrating $f (x) g (x) = x^{2} \sin x$ $f(x)g(x) = x^2 sinx$ is to be worked out.

Knowing that integration is anti-derivative, let us examine how product of sub-functions is differentiated.
$(u v)' = v u' + u v'$ $(uv)′ = vu′+uv′$

If one of the functions is in $x^{n}$ $x^n$ form, then repeated differentiation will result in a constant. Knowing this, the equation above is modified to
$u v' = (u v)' - v u'$ $u v′ = (uv)′ - vu′$

integrating this
$\int u v' d x = \int (u v)' d x - \int v u' d x$ $int u v′ dx = int (uv)′ dx - int v u′ dx$

$\int u v' d x = u v - \int v u' d x$ $int u v′ dx = uv - int v u′ dx$

substituting $u = f (x)$ $u= f(x)$ and $v' = g (x)$ $v′ = g(x)$ and so $v = \int g (x) d x$ $v= int g(x) dx$

$\int f (x) g (x) d x$ $int f(x) g(x) dx$ $= f (x) \int g (x) d x$ $= f(x) intg(x)dx$ $- \int [\int g (x) d x] [\frac{d}{d x} f (x)] d x$ $- int [int g(x) dx] [d/(dx)f(x)] dx$

The advantage is, the function $f (x)$ $f(x)$ is differentiated in the second term. If $f (x) = x$ $f(x)=x$ then the second term is simplified to a standard result.

$\int f (x) g (x) d x$ $int f(x) g(x) dx$ $= f (x) \int g (x) d x$ $= f(x) intg(x)dx$ $- \int [\int g (x) d x] [\frac{d}{d x} f (x)] d x$ $- int [int g(x) dx] [d/(dx)f(x)] dx$

This is called integration by parts. Note that the integrand is split into two parts.

example

$\int x^{2} \sin x d x$ $int x^2 sin x dx$

using integration by parts
$= - x^{2} \cos x - \int (- \cos x) (2 x) d x + c$ $=-x^2cos x - int (-cos x) (2x)dx+c$

$= - x^{2} \cos x + \int 2 x \cos x d x + c$ $=-x^2cos x + int 2xcos x dx+c$

The integration by parts can be used on the second term too.

$= - x^{2} \cos x + \int 2 x \cos x d x + c$ $=-x^2cos x + int 2xcos x dx+c$

using integration by parts on the second term
$= - x^{2} \cos x + 2 x \sin x - \int \sin x \times 2 d x + c$ $=-x^2cos x + 2x sin x - int sinx xx2 dx+ c$

$= - x^{2} \cos x + 2 x \sin x + 2 \cos x + c$ $= -x^2cos x + 2x sin x + 2cos x+ c$

interesting form with e

Interesting observation in the integration by parts is that one of the two sub-functions is differentiated sequentially. The other is integrated repeatedly. If one of the functions is $e^{x}$ $e^x$ , then as the sequential integration or differentiation does not change $e^{x}$ $e^x$ . This leads to an interesting form of integration.

$\int e^{x} \sin x d x$ $int e^x sin x dx$

$= e^{x} (- \cos x)$ $= e^x(-cos x)$ $+ \int e^{x} \cos x d x + c$ $+ int e^x cos x dx + c$

$= - e^{x} \cos x + e^{x} \sin x$ $= -e^xcos x + e^x sinx$ $- \int e^{x} \sin x d x + c$ $- int e^x sin x dx + c$

The result has the integral we started off with. We can handle this as an equation and rearrange such that the term is in one side of the equation".

$I = \int e^{x} \sin x d x$ $I = int e^x sin x dx$

$= e^{x} (- \cos x)$ $= e^x(-cos x)$ $+ \int e^{x} \cos x d x + c$ $+ int e^x cos x dx + c$

$= - e^{x} \cos x + e^{x} \sin x$ $= -e^xcos x + e^x sinx$ $- \int e^{x} \sin x d x + c$ $- int e^x sin x dx + c$

The above is written as $I = - e^{x} \cos x + e^{x} \sin x - I + c$ $I = -e^xcos x + e^x sinx - I + c$

$I = (- e^{x} \cos x + e^{x} \sin x) / 2 + c$ $I=(-e^xcos x + e^x sinx)//2 + c$

interesting form with h + h'

Another interesting observation in integration by parts: If one of the functions is $e^{x}$ $e^x$ , then as the sequential integration or differentiation does not change $e^{x}$ $e^x$ . This leads to an interesting form of integration when the other function is in the form. $h (x) + h' (x)$ $h(x)+h′(x)$ .

$\int e^{x} (h (x) + h' (x)) d x$ $int e^x (h(x)+h′(x)) dx$

$= \int e^{x} h (x) d x$ $= int e^x h(x) dx$ $+ \int e^{x} h' (x) d x$ $+ int e^x h′(x) dx$

applying integration by parts in the first integral

$= e^{x} h (x)$ $= e^x h(x)$ $- \int e^{x} h' (x) d x + \int e^{x} h' (x) d x + c$ $- int e^x h′(x) dx + int e^x h′(x) dx +c$

cancel the second term and third term. The integration result is $e^{x} h (x) + c$ $e^xh(x) + c$ .

Integrate $\int e^{x} (\ln x + \frac{1}{x}) d x$ $int e^x (ln x + 1/x) dx$ .

$\int e^{x} (\ln x + \frac{1}{x}) d x$ $int e^x (ln x + 1/x) dx$

$= \int e^{x} \ln x d x + \int e^{x} \frac{1}{x} d x$ $= int e^x ln x dx + int e^x 1/x dx$ $= e^{x} \ln x$ $= e^x ln x$ $- \int e^{x} \frac{1}{x} d x + \int e^{x} \frac{1}{x} d x + c$ $- int e^x 1/x dx + int e^x 1/x dx +c$ $= e^{x} \ln x + c$ $= e^x ln x +c$

summary

Integration by Parts: $\int f (x) g (x) d x$ $int f(x) g(x) dx$ $= f (x) \int g (x) d x - \int [\int g (x) d x] [\frac{d}{d x} f (x)] d x$ $= f(x) int g(x)dx - int [int g(x) dx] [d/(dx)f(x)] dx$

$e^{x} \sin x$ $e^x sin x$ or $e^{x} \cos x$ $e^x cos x$ in Integration by Parts: After two iterations of integration by parts, the integral is solved by rearranging the terms of the equation.

$e^{x}$ $e^x$ in Integration by Parts: $\int e^{x} (h (x) + h' (x)) d x$ $int e^x (h(x)+h′(x)) dx$ $= e^{x} h (x) + c$ $=e^x h(x) + c$

examples

Integrate $\int e^{x} \arctan x + \frac{e^{x}}{1 + x^{2}} d x$ $int e^x arctan x + e^x/(1+x^2) dx$

The answer is $e^{x} \arctan x + c$ $e^x arctan x + c$ ". The integral is in the form $\int e^{x} (h (x) + h' (x)) d x$ $int e^x (h(x)+h′(x)) dx$

Integrate $\int e^{x} \cos x d x$ $int e^x cos x dx$

The answer is " $e^{x} (\sin x + \cos x) / 2 + c$ $e^x (sin x+cos x)//2 + c$ ".

Integrate $\int x \sin x d x$ $int x sin x dx$

The answer is " $- x \cos x + \sin x + c$ $-x cos x + sinx +c$ ".

summary

Outline

The outline of material to learn "Integral Calculus" is as follows.

• Detailed outline of Integral Calculus

→ Application Scenario

→ Integration First Principles

→ Graphical Meaning of Integration

→ Definition of Integrals

→ Fundamental Theorem of Calculus

→ Algebra of Integrals

→ Antiderivatives: Standard results

→ Integration of Expressions

→ Integration by Substitution

→ Integration using Identities

→ Integration by Parts

→ Integration by Partial Fraction

→ Integration: Combination of Methods