maths > integral-calculus

Integration by Parts

what you'll learn...

Overview

Integration by Parts :

»  $\int f\left(x\right)g\left(x\right)dx$$\int f \left(x\right) g \left(x\right) \mathrm{dx}$ $=f\left(x\right)\int g\left(x\right)dx$$= f \left(x\right) \int g \left(x\right) \mathrm{dx}$$-\int \left[\int g\left(x\right)dx\right]\left[\frac{d}{dx}f\left(x\right)\right]dx$$- \int \left[\int g \left(x\right) \mathrm{dx}\right] \left[\frac{d}{\mathrm{dx}} f \left(x\right)\right] \mathrm{dx}$

→  $\int {e}^{x}\left(h\left(x\right)+h\prime \left(x\right)\right)dx$$={e}^{x}h\left(x\right)+c$$= {e}^{x} h \left(x\right) + c$

part by part

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

Knowing that integration is anti-derivative, let us examine how product of sub-functions is differentiated.
$\left(uv\right)\prime =vu\prime +uv\prime$

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
$uv\prime =\left(uv\right)\prime -vu\prime$

integrating this
$\int uv\prime dx=\int \left(uv\right)\prime dx-\int vu\prime dx$

$\int uv\prime dx=uv-\int vu\prime dx$

substituting $u=f\left(x\right)$$u = f \left(x\right)$ and $v\prime =g\left(x\right)$ and so $v=\int g\left(x\right)dx$$v = \int g \left(x\right) \mathrm{dx}$

$\int f\left(x\right)g\left(x\right)dx$$\int f \left(x\right) g \left(x\right) \mathrm{dx}$ $=f\left(x\right)\int g\left(x\right)dx$$= f \left(x\right) \int g \left(x\right) \mathrm{dx}$$-\int \left[\int g\left(x\right)dx\right]\left[\frac{d}{dx}f\left(x\right)\right]dx$$- \int \left[\int g \left(x\right) \mathrm{dx}\right] \left[\frac{d}{\mathrm{dx}} f \left(x\right)\right] \mathrm{dx}$

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

$\int f\left(x\right)g\left(x\right)dx$$\int f \left(x\right) g \left(x\right) \mathrm{dx}$ $=f\left(x\right)\int g\left(x\right)dx$$= f \left(x\right) \int g \left(x\right) \mathrm{dx}$$-\int \left[\int g\left(x\right)dx\right]\left[\frac{d}{dx}f\left(x\right)\right]dx$$- \int \left[\int g \left(x\right) \mathrm{dx}\right] \left[\frac{d}{\mathrm{dx}} f \left(x\right)\right] \mathrm{dx}$

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

example

$\int {x}^{2}\mathrm{sin}xdx$$\int {x}^{2} \sin x \mathrm{dx}$

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

$=-{x}^{2}\mathrm{cos}x+\int 2x\mathrm{cos}xdx+c$$= - {x}^{2} \cos x + \int 2 x \cos x \mathrm{dx} + c$

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

$=-{x}^{2}\mathrm{cos}x+\int 2x\mathrm{cos}xdx+c$$= - {x}^{2} \cos x + \int 2 x \cos x \mathrm{dx} + c$

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

$=-{x}^{2}\mathrm{cos}x+2x\mathrm{sin}x+2\mathrm{cos}x+c$$= - {x}^{2} \cos x + 2 x \sin x + 2 \cos 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}\mathrm{sin}xdx$$\int {e}^{x} \sin x \mathrm{dx}$

$={e}^{x}\left(-\mathrm{cos}x\right)$$= {e}^{x} \left(- \cos x\right)$$+\int {e}^{x}\mathrm{cos}xdx+c$$+ \int {e}^{x} \cos x \mathrm{dx} + c$

$=-{e}^{x}\mathrm{cos}x+{e}^{x}\mathrm{sin}x$$= - {e}^{x} \cos x + {e}^{x} \sin x$$-\int {e}^{x}\mathrm{sin}xdx+c$$- \int {e}^{x} \sin x \mathrm{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}\mathrm{sin}xdx$$I = \int {e}^{x} \sin x \mathrm{dx}$

$={e}^{x}\left(-\mathrm{cos}x\right)$$= {e}^{x} \left(- \cos x\right)$$+\int {e}^{x}\mathrm{cos}xdx+c$$+ \int {e}^{x} \cos x \mathrm{dx} + c$

$=-{e}^{x}\mathrm{cos}x+{e}^{x}\mathrm{sin}x$$= - {e}^{x} \cos x + {e}^{x} \sin x$$-\int {e}^{x}\mathrm{sin}xdx+c$$- \int {e}^{x} \sin x \mathrm{dx} + c$

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

$I=\left(-{e}^{x}\mathrm{cos}x+{e}^{x}\mathrm{sin}x\right)/2+c$$I = \left(- {e}^{x} \cos x + {e}^{x} \sin x\right) / 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. $h\left(x\right)+h\prime \left(x\right)$.

$\int {e}^{x}\left(h\left(x\right)+h\prime \left(x\right)\right)dx$

$=\int {e}^{x}h\left(x\right)dx$$= \int {e}^{x} h \left(x\right) \mathrm{dx}$$+\int {e}^{x}h\prime \left(x\right)dx$

applying integration by parts in the first integral

$={e}^{x}h\left(x\right)$$= {e}^{x} h \left(x\right)$$-\int {e}^{x}h\prime \left(x\right)dx+\int {e}^{x}h\prime \left(x\right)dx+c$

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

Integrate $\int {e}^{x}\left(\mathrm{ln}x+\frac{1}{x}\right)dx$$\int {e}^{x} \left(\ln x + \frac{1}{x}\right) \mathrm{dx}$.

$\int {e}^{x}\left(\mathrm{ln}x+\frac{1}{x}\right)dx$$\int {e}^{x} \left(\ln x + \frac{1}{x}\right) \mathrm{dx}$

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

summary

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

${e}^{x}\mathrm{sin}x$${e}^{x} \sin x$ or ${e}^{x}\mathrm{cos}x$${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}$${e}^{x}$ in Integration by Parts: $\int {e}^{x}\left(h\left(x\right)+h\prime \left(x\right)\right)dx$$={e}^{x}h\left(x\right)+c$$= {e}^{x} h \left(x\right) + c$

examples

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

The answer is ${e}^{x}\mathrm{arctan}x+c$${e}^{x} \arctan x + c$". The integral is in the form $\int {e}^{x}\left(h\left(x\right)+h\prime \left(x\right)\right)dx$

Integrate $\int {e}^{x}\mathrm{cos}xdx$$\int {e}^{x} \cos x \mathrm{dx}$

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

Integrate $\int x\mathrm{sin}xdx$$\int x \sin x \mathrm{dx}$

The answer is "$-x\mathrm{cos}x+\mathrm{sin}x+c$$- x \cos x + \sin x + c$".

summary

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

${e}^{x}\mathrm{sin}x$${e}^{x} \sin x$ or ${e}^{x}\mathrm{cos}x$${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}$${e}^{x}$ in Integration by Parts: $\int {e}^{x}\left(h\left(x\right)+h\prime \left(x\right)\right)dx$$={e}^{x}h\left(x\right)+c$$= {e}^{x} h \left(x\right) + c$

Outline