## 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

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

$$u= 2x-1$$

$$du= 2dx$$

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

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

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

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

#### 8 comentarios:

1. I don't like maths =P

2. My god the equations! :)

3. What program do you use to write those equations down?

4. I used a script called mathjax. If you whant to use it tell and I show you how

5. Why? Why would anyone make a blog talking about integrals? I don't suppose you'll be doing Taylor series and 3d integrals now, will ya?

6. i will try to solve!

7. Let's see what he will be posting in the future :P

8. I'm so bad in math xD