Integral Calculus : Integration by Substitution

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Integration by Substitution

what you'll learn...

Overview

Integration by Substitution :

When $f (x) = h (g (x)) g' (x)$ $f(x) = h(g(x)) g′(x)$

then,
$\int f (x) d x = \int h (y) d y$ $int f(x)dx = int h(y)dy$
where $y = g (x)$ $y=g(x)$
and $d y = g' (x) d x$ $dy = g′(x)dx$ and so
» $\int f (x) d x$ $int f(x)dx$ $= [\int h (y) d y] ∣_{y = g (x)} + c$ $= [int h(y)dy]|_(y=g(x)) + c$

substitute

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

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

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

Substitute $t = x^{2}$ $t=x^2$ and $d t = 2 x d x$ $dt = 2xdx$ .

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

$= \int \sin t d t$ $= int sin t dt$

$= - \cos t + c$ $= -cos t +c$

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

Considering the integration $\int f (x) d x$ $int f(x)dx$ :
If it is observed that $f (x)$ $f(x)$ can be rewritten as
$f (x) = h [g (x)] g' (x)$ $f(x) = h[g(x)] g′(x)$

then,
$\int f (x) d x = \int h (y) d y$ $int f(x)dx = int h(y)dy$
where $y = g (x)$ $y=g(x)$
and $d y = g' (x) d x$ $dy = g′(x)dx$

This method is "integration by substitution".

summary

Integration by Substitution: When $f (x) = h (g (x)) g' (x)$ $f(x) = h(g(x)) g′(x)$

then,
$\int f (x) d x = \int h (y) d y$ $int f(x)dx = int h(y)dy$
where $y = g (x)$ $y=g(x)$
and $d y = g' (x) d x$ $dy = g′(x)dx$ and so $\int f (x) d x$ $int f(x)dx$ $= [\int h (y) d y] ∣_{y = g (x)} + c$ $= [int h(y)dy]|_(y=g(x)) + c$

parametric form

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

summary

Given a function in parametric form $y = r \cos t$ $y=rcos t$ and $x = r \sin t$ $x=rsin t$ .
$\int y d x$ $int ydx$

substituting $y = r \cos t$ $y=rcos t$ and $d x = d (r \sin t)$ $dx = d(r sin t)$ . the $d ()$ $d()$ means derivative of
$= \int r \cos t d (r \sin t)$ $= int r cost d(r sin t)$

$= \int r \cos t r \cos t d t$ $= int r cos t r cos t dt$

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

This can be integrated.

tan

Integrate $\int \tan x d x$ $int tan x dx$

This can be integrated with $\tan x = - (\cos x)' / \cos x$ $tan x = - (cos x)′//cos x$

$\int \tan x d x$ $int tanx dx$

$= \int \frac{- (\cos x)'}{\cos x}$ $=int (- (cos x)′)/(cos x)$

$= - \log | \cos x | + c$ $=-log|cos x| + c$

cot

Integrate $\int \cot x d x$ $int cot x dx$

This can be integrated with $\cot x = (\sin x)' / \sin x$ $cot x = (sin x)′//sin x$

$\int \cot x d x$ $int cotx dx$

$= \frac{(\sin x)'}{\sin x}$ $=((sin x)′)/(sin x)$

$= \log | \sin x | + c$ $=log|sin x| + c$

sec

Integrate $\int \sec x d x$ $int sec x dx$

This can be integrated with $\sec x$ $sec x$ $= (\sec x + \tan x)'$ $= (sec x + tan x)′$ $/ (\sec x + \tan x)$ $//(sec x + tan x)$

$\int \sec x d x$ $int sec x dx$

$= \int \frac{(\sec x + \tan x)'}{\sec x + \tan x} d x$ $=int ((sec x + tan x)′)/(sec x + tan x) dx$

$= \log | \sec x + \tan x | + c$ $=log|sec x + tan x|+c$

cosec

Integrate $\int \csc x d x$ $int csc x dx$

This can be integrated with $\csc x$ $csc x$ $= - (\csc x + \cot x)'$ $= -(csc x + cot x)′$ $/ (\csc x + \cot x)$ $//(csc x + cot x)$

$\int \csc x d x$ $int csc x dx$

$= \int \frac{- (\csc x + \cot x)'}{\csc x + \cot x} d x$ $=int (-(csc x + cot x)′)/(csc x + cot x) dx$

$= - \log | \csc x + \cot x | + c$ $=-log|cscx+ cotx| + c$

summary

Integration by Substitution

When $f (x) = h (g (x)) g' (x)$ $f(x) = h(g(x)) g′(x)$

then,
$\int f (x) d x = \int h (y) d y$ $int f(x)dx = int h(y)dy$
where $y = g (x)$ $y=g(x)$
and $d y = g' (x) d x$ $dy = g′(x)dx$ and so
» $\int f (x) d x$ $int f(x)dx$ $= [\int h (y) d y] ∣_{y = g (x)} + c$ $= [int h(y)dy]|_(y=g(x)) + c$

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