# Integral Substitution

• Aug 22nd 2010, 02:21 PM
jjstuart
Integral Substitution
Hi,
Im trying to teach myself the substitution rule for integrals and am having a problem with the formula. I'm not grasping something here and hope someone can help me out.

Here is the integral I'm working with.

4x^3 * sqrt(x^4 + 7) dx

The formula says f(g(x)g'(x)dx = Integral f(u) du

To me that says to take f(u) * du, but when they work the problem out it seems that they just drop the du.

If we start out with the integral

4x^3 * sqrt(x^4 + 7) dx

g(x) = u
u = x^4 + 7
du = 4x^3

Now if I plug in the numbers following the formula I would write this out.

(x^4 + 7)^1/2 * (4x^3)

Then the examples goes on to say for the next step you end up with this

2/3(u)^3/2 + c

What happens to the 4x^3? From reading the formula I would think we have to multiple u and du? Any help is appreciated.

Thanks
• Aug 22nd 2010, 02:30 PM
skeeter
Quote:

Originally Posted by jjstuart
Hi,
Im trying to teach myself the substitution rule for integrals and am having a problem with the formula. I'm not grasping something here and hope someone can help me out.

Here is the integral I'm working with.

4x^3 * sqrt(x^4 + 7) dx

The formula says f(g(x)g'(x)dx = Integral f(u) du

To me that says to take f(u) * du, but when they work the problem out it seems that they just drop the du.

If we start out with the integral

4x^3 * sqrt(x^4 + 7) dx

g(x) = u
u = x^4 + 7
du = 4x^3 dx

note the correction above ...

$\displaystyle \frac{du}{dx} = 4x^3$

$du = 4x^3 \, dx$
• Aug 22nd 2010, 02:39 PM
Vlasev
You just need to write it out more and you'll get it! Here

$\displaystyle u = x^4 + 7$
$\displaystyle du = 4x^3 dx$

Thus $\displaystyle dx = \frac{du}{4x^3}$

Now, replace these in the integral to get

$\displaystyle \int 4x^3\sqrt{x^4 + 7} \;dx = \int 4x^3\sqrt{u} \frac{du}{4x^3} = \int \sqrt{u}\;du$

Since the $4x^3$ factor cancels out!
• Aug 22nd 2010, 04:17 PM
jjstuart
Thank you for the replies. I see how they cancel out now. I guess the hardest part for me too understand is how dx = du/4x3. I think I have more reading up to do on change of variables, I think?
• Aug 22nd 2010, 04:42 PM
yeKciM

Before continuing learn table integrals :D (about 20-30 depending where someone is considering some of them to be or not to be table integrals) :D

1° if you have function $f(x)$ so that

$\displaystyle \int f(x) \;dx = \int f(\varphi (t)) \varphi ' (t) \; dt$

where $x= \varphi (t)$

it's basic way to solve it with substitution... (by which you will reduce your integral to some of the basic "table" integrals ) Use suitable substitution $x=\varphi (t)$ of the new variable "t", you have to express everything with "t" and of course $dx=\varphi ' (t) \;dt$ also :D if you don't succeed in that than you didn't chose good substitution :D (that is not a big problem... just go back and chose another one :D:D:D:D that all comes from much much practice :D)

let's look few examples (simple ones... from those you start :D )

1) example

$I=\int (x+1)^{100} \;dx$

substitute :

$(x+1) = t$

$dx=dt$

$I=\int (x+1)^{100} \;dx = \int t^{100} \; dt = \frac {t^{101}}{101} + C = \frac {(x+1)^{101}}{101} +C$

and much more of those but point is that integrals shaped like this :

$\int f(ax+b) \;dx$

using substitute $(ax+b)=t$ you can simplify your integral so that :

$\int f(ax+b) \;dx =\frac {1}{a} \int f(t) \; dt$

]integrals shaped like this :

$I= \int \frac {dx}{ax\pm b}$

using substitute

$(ax\pm b)=t$

$dx=\frac{dt}{a}$

you can simplify your integral so that:

$I= \int \frac {dx}{ax\pm b}= \int \frac {\frac {dt}{a}}{t}=\frac {1}{a} \int \frac {dt}{t}=\frac {1}{a} ln |t| + C = \frac {1}{a} ln|ax\pm b| +C$

2) example

$I= \int \frac {dx}{1+4x^2} = \int \frac {dx}{1+(2x)^2}$

substitute

$2x=t \Rightarrow 2dx=dt \rightarrow dx=\frac{dt}{2}$

$I= \int \frac {dx}{1+4x^2} = . . . . . = \frac {1}{2} \arctan {2x} + C$

3) example

$I= \int \frac {dx}{a^2+x^2 }$

this one is table one but... :D:D:D

$I= \int \frac {dx}{a^2+x^2 } = \int \frac {dx}{a^2(1+(\frac {x}{a})^2) } = \frac {1}{a^2} \int \frac {dx}{1+(\frac {x}{a})^2}$

substitute $\frac {x}{a}=t \Rightarrow dx=a \; dt$

$I= \int \frac {dx}{a^2+x^2 }= . . . .=\frac {1}{a} \arctan{\frac {x}{a}} +C$

4) example

$I = \int \frac {dx}{\sqrt{x^2\pm a^2}}$

this one can be used as table integral ...

substitute

$\sqrt{x^2\pm a^2}=t-x \Rightarrow \sqrt{x^2\pm a^2}+x=t$

$( \frac {x}{\sqrt{x^2\pm a^2} +1} )dx=dt \Rightarrow . . . \Rightarrow dx=\frac {t-x}{t} dt$

$I = \int \frac {dx}{\sqrt{x^2\pm a^2}}= \int \frac {\frac {t-x}{t}}{t-x}\; dt = \int \frac {dt}{t} = \ln {t} +C = \ln {\sqrt{x^2\pm a^2}+x} +C$

anyway, there is much to be said about integration :D and methods of solving integrals. These up there are on substitution based :D Perhaps (sure) i miss few key examples (sorry about that, I will edit if it comes to my mind) But for get along with integrals you should practice (learn basic/table integrals, because you when solving integrals you are just finding the way to write your integral similar to any of those basic) substitution, partial integration... and to know basic types (because any integral is solved by the pattern :D:D:D:D)

those are :

$\displaystyle I= \int \frac {P(x)}{Q(x)} \;dx$

$\displaystyle I = \int R(x,\sqrt [n]{ax+b}) \;dx$

$\displaystyle I= \int R(x,\sqrt [n]{\frac {ax+b}{cx+d}} )\;dx$

$\displaystyle I= \int x^p (ax^q+b)^r \;dx$

$\displaystyle I= \int \frac {dx}{ax^2+bx+c} \;dx$

$\displaystyle I= \int \frac {dx}{\sqrt{ax^2+bx+c}} \;dx$

$\displaystyle I= \int \sqrt{ax^2+bx+c} \;dx$

$\displaystyle I= \int \frac {P_n(x)}{\sqrt{ax^2+bx+c}} \;dx$

$\displaystyle I= \int P_n(x) \sqrt{ax^2+bx+c} \;dx$

$\displaystyle I= \int R(\sin {x} , \cos {x} ) \; dx$

and so on... variation of those and perhaps few that I forgot to put them there :D but like i said, this to get you started :D:D:D:D good luck :D

P.S. just practice and practice, that will make difference from good substitutions and bad ones :D:D:D:D
• Aug 22nd 2010, 06:09 PM
jjstuart
Thank you! It makes a lot more sense to me now. You're right, I just need to sit down and work out a lot more problems.