 maths > integral-calculus

Integration by Substitution

what you'll learn...

Overview

Integration by Substitution :

When $f\left(x\right)=h\left(g\left(x\right)\right)g\prime \left(x\right)$

then,
$\int f\left(x\right)dx=\int h\left(y\right)dy$$\int f \left(x\right) \mathrm{dx} = \int h \left(y\right) \mathrm{dy}$
where $y=g\left(x\right)$$y = g \left(x\right)$
and $dy=g\prime \left(x\right)dx$ and so
»  $\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$$=\left[\int h\left(y\right)dy\right]{\mid }_{y=g\left(x\right)}+c$$= \left[\int h \left(y\right) \mathrm{dy}\right] {|}_{y = g \left(x\right)} + c$

substitute

Consider the integration $\int 2x{\mathrm{sin}x}^{2}dx$$\int 2 x \sin {x}^{2} \mathrm{dx}$
We notice that $\frac{d}{dx}{x}^{2}=2x$$\frac{d}{\mathrm{dx}} {x}^{2} = 2 x$ is part of the integration.

$t={x}^{2}$$t = {x}^{2}$ and substitute in the integrand.

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

Substitute $t={x}^{2}$$t = {x}^{2}$ and $dt=2xdx$$\mathrm{dt} = 2 x \mathrm{dx}$.

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

$=\int \mathrm{sin}tdt$$= \int \sin t \mathrm{dt}$

$=-\mathrm{cos}t+c$$= - \cos t + c$

substitute $t={x}^{2}$$t = {x}^{2}$
$=-{\mathrm{cos}x}^{2}+c$$= - \cos {x}^{2} + c$

Considering the integration $\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$ :
If it is observed that $f\left(x\right)$$f \left(x\right)$ can be rewritten as
$f\left(x\right)=h\left[g\left(x\right)\right]g\prime \left(x\right)$

then,
$\int f\left(x\right)dx=\int h\left(y\right)dy$$\int f \left(x\right) \mathrm{dx} = \int h \left(y\right) \mathrm{dy}$
where $y=g\left(x\right)$$y = g \left(x\right)$
and $dy=g\prime \left(x\right)dx$

This method is "integration by substitution".

summary

Integration by Substitution: When $f\left(x\right)=h\left(g\left(x\right)\right)g\prime \left(x\right)$

then,
$\int f\left(x\right)dx=\int h\left(y\right)dy$$\int f \left(x\right) \mathrm{dx} = \int h \left(y\right) \mathrm{dy}$
where $y=g\left(x\right)$$y = g \left(x\right)$
and $dy=g\prime \left(x\right)dx$ and so $\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$$=\left[\int h\left(y\right)dy\right]{\mid }_{y=g\left(x\right)}+c$$= \left[\int h \left(y\right) \mathrm{dy}\right] {|}_{y = g \left(x\right)} + c$

parametric form

Given a function in parametric form $y=r\mathrm{cos}t$$y = r \cos t$ and $x=r\mathrm{sin}t$$x = r \sin t$, To proceed with the integration, substitute $y=r\mathrm{cos}t$$y = r \cos t$ and $dx=d\left(r\mathrm{cos}t\right)$$\mathrm{dx} = d \left(r \cos t\right)$. Then, proceed with variable of integration as $t$$t$

summary

Given a function in parametric form $y=r\mathrm{cos}t$$y = r \cos t$ and $x=r\mathrm{sin}t$$x = r \sin t$.
$\int ydx$$\int y \mathrm{dx}$

substituting $y=r\mathrm{cos}t$$y = r \cos t$ and $dx=d\left(r\mathrm{sin}t\right)$$\mathrm{dx} = d \left(r \sin t\right)$. the $d\left(\right)$$d \left(\right)$ means derivative of
$=\int r\mathrm{cos}td\left(r\mathrm{sin}t\right)$$= \int r \cos t d \left(r \sin t\right)$

$=\int r\mathrm{cos}tr\mathrm{cos}tdt$$= \int r \cos t r \cos t \mathrm{dt}$

$=\int {r}^{2}{\mathrm{cos}}^{2}tdt$$= \int {r}^{2} {\cos}^{2} t \mathrm{dt}$

This can be integrated.

tan

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

This can be integrated with $\mathrm{tan}x=-\left(\mathrm{cos}x\right)\prime /\mathrm{cos}x$

$\int \mathrm{tan}xdx$$\int \tan x \mathrm{dx}$

$=\int \frac{-\left(\mathrm{cos}x\right)\prime }{\mathrm{cos}x}$

$=-\mathrm{log}|\mathrm{cos}x|+c$$= - \log | \cos x | + c$

cot

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

This can be integrated with $\mathrm{cot}x=\left(\mathrm{sin}x\right)\prime /\mathrm{sin}x$

$\int \mathrm{cot}xdx$$\int \cot x \mathrm{dx}$

$=\frac{\left(\mathrm{sin}x\right)\prime }{\mathrm{sin}x}$

$=\mathrm{log}|\mathrm{sin}x|+c$$= \log | \sin x | + c$

sec

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

This can be integrated with $\mathrm{sec}x$$\sec x$$=\left(\mathrm{sec}x+\mathrm{tan}x\right)\prime$$/\left(\mathrm{sec}x+\mathrm{tan}x\right)$$/ \left(\sec x + \tan x\right)$

$\int \mathrm{sec}xdx$$\int \sec x \mathrm{dx}$

$=\int \frac{\left(\mathrm{sec}x+\mathrm{tan}x\right)\prime }{\mathrm{sec}x+\mathrm{tan}x}dx$

$=\mathrm{log}|\mathrm{sec}x+\mathrm{tan}x|+c$$= \log | \sec x + \tan x | + c$

cosec

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

This can be integrated with $\mathrm{csc}x$$\csc x$$=-\left(\mathrm{csc}x+\mathrm{cot}x\right)\prime$$/\left(\mathrm{csc}x+\mathrm{cot}x\right)$$/ \left(\csc x + \cot x\right)$

$\int \mathrm{csc}xdx$$\int \csc x \mathrm{dx}$

$=\int \frac{-\left(\mathrm{csc}x+\mathrm{cot}x\right)\prime }{\mathrm{csc}x+\mathrm{cot}x}dx$

$=-\mathrm{log}|\mathrm{csc}x+\mathrm{cot}x|+c$$= - \log | \csc x + \cot x | + c$

summary

Integration by Substitution

When $f\left(x\right)=h\left(g\left(x\right)\right)g\prime \left(x\right)$

then,
$\int f\left(x\right)dx=\int h\left(y\right)dy$$\int f \left(x\right) \mathrm{dx} = \int h \left(y\right) \mathrm{dy}$
where $y=g\left(x\right)$$y = g \left(x\right)$
and $dy=g\prime \left(x\right)dx$ and so
»  $\int f\left(x\right)dx$$\int f \left(x\right) \mathrm{dx}$$=\left[\int h\left(y\right)dy\right]{\mid }_{y=g\left(x\right)}+c$$= \left[\int h \left(y\right) \mathrm{dy}\right] {|}_{y = g \left(x\right)} + c$

Outline