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lunes, 27 de junio de 2011

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lunes, 9 de mayo de 2011

More Examples of Integral by Substitution

#1

$\int\frac{tan^{-1}x}{1+x^{2}}\;dx$

$u=tan^{-1}x$
$du=\frac{dx}{1+x^2}$
$\int u\;du=\frac{u^2}{2}+C$
$\frac{(tan^{-1}x)^{2}{}}{2}+C$

#2

$\int sinx*cosx\;dx$
$u= sinx$
$du= cosxdx$
$\int u\;du= \frac{1}{2} u^2+c$
$\frac{1}{2} sin^2x+c$

#3

$\int sin^2 3x\cos 3x\;dx$
$u= 3x$
$du= 3dx$
$\frac{1}{3}du= dx$
$\frac{1}{3}\int sin^2 u\cos u\;du$
$u= sin u$
$du= cos u\;du$
$\frac{1}{3}\int u^2\;du = \frac{1}{3}\;(\frac{1}{3}u^3)+c$
$\frac{1}{9}u^3+c$
$\frac{1}{9}sin^3\;3x+c$

#4

$\int e^x(1+e^x)^{10}\;dx$
$u= 1+e^x$
$du= e^x\;dx$
$\int u^{10}\;du$
$=\frac{u^{11}}{11}$
$=\frac{(1+e^x)^{11}}{11}+C$

#5

$\int \frac{(lnx)^2}{x}\;dx$
$u= lnx$
$du= \frac{1}{x}\;dx$
$\int u^2\;du$
$=\frac{(u)^3}{3}$
$=\frac{(lnx)^3}{3}+C$

#6

$$\int \frac{1}{2x-1}\;dx$$

$$u= 2x-1$$

$$du= 2dx$$

$$\frac{1}{2}du=dx$$

$$\frac{1}{2}\int \frac{1}{u}\;du$$

$$=\frac{1}{2}lnu+C$$

$$=\frac{1}{2}ln(2x-1)+C$$

Integral by Substitution Example #4

$\int sinx*cosx\;dx$

$u= sinx$

$du= cosx \; dx$

$ \int u\;du= \frac{1}{2} u^2+c $

$ \frac{1}{2} sin^2x+c $

Integral by Substitution Example #3

Calculate
$\int_0^{\sqrt{\pi}}x\cos (x^2)\;dx$

SOLUTION
Let
$\begin{align*} u &= x^2 \\ du &= 2\,dx \end{align*}$

To find the limits of integration we note that when,

$x=0 \rightarrow u = 0^2 = 0$

and when

$x=\sqrt{\pi} \rightarrow u = \sqrt{\pi}^{\;2} = \pi$

Therefore
$\begin{align*} \int_0^{\sqrt{\pi}}x\cos (x^2)\;dx&=\frac{1}{2}\int_0^{\pi}\cos (u)\;du \\ &= \left. \frac{1}{2} \sin(u) \right |_0^\pi\\ &= \frac{1}{2} \left ( \sin(\pi)-\sin (0) \right ) \\ &= 0 \end{align*}$

Integral by Substitution Example #2

Calculate
$\int \frac{1+4x}{\sqrt{1+x+2x^2}}dx$

SOLUTION
Let
u= 1+x+2x^2

then

therefore
$\begin{align*} \int \frac{1+4x}{\sqrt{1+x+2x^2}}dx &= \int \frac{du}{u^{\frac{1}{2}}} \\ &= \int u^{-\frac{1}{2}}du \\ &= 2u^{\frac{1}{2}}+C \\ &= 2\sqrt{1+x+2x^2}+C\;\;\;\square \end{align*}$

viernes, 6 de mayo de 2011

Integration by Sustitution

Lets start by understanding that this method of integration comes from the rule chain method for example,

$\frac{d[f[g(x)]}{dx}=f'[g(x)]g'(x)$ (1)

lets integrate both sides of equation 1

$\begin{align*} \int \frac{d[f[g(x)]}{dx}\; dx&=\int f'[g(x)]g'(x)\;dx \\ f[g(x)]&= \int f'[g(x)]g'(x)\;dx \\ \end{align*}$

$\int f'[g(x)]g'(x)\;dx$

we can use the substitution

working with

u=g(x)

(2)

then

$du=g'(x)\;dx$

we make the substitution on the integral and get,

$\int f'(u) du$

by the fundamental theorem of calcules we get

$\int f'(u) du=f(u)$

we use equation (2) and get that

$\int f'[g(x)]g'(x)dx=f[g(x)]+c$

Example #1

Solve the integral

$\int (x^2+1)^{20} \; dx$

We choose

u=x^2+1

and

$du=2x\;dx$

we change the integral to

$\int u^{20} \; \frac{dx}{2}=\frac{1}{2}\int u^{20}\;du=\frac{1}{2}\frac{1}{21}u^{21}=\frac{1}{42}u^{21}$

My first post!

For this blog im going to use latex. You can learn a lot by using this page http://www.codecogs.com/latex/eqneditor.php

Hope that the problems and solutions I post help you get better understanding of school and university math.

Bye