How to integrate a function with a square root in it


by rock.freak667
Tags: function, integrate, root, square
rock.freak667
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#1
Sep3-07, 05:27 PM
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1. The problem statement, all variables and given/known data

find [tex] \int x^{\frac{3}{2}}\sqrt{1+x} dx [/tex]

2. Relevant equations



3. The attempt at a solution

Now i tried all the methods i know of which include integration by parts and substitution,the integration by parts didn't work too well, so substitution I tried.
let [tex]u=tan^2x[/tex] and [tex]du = 2tanxsec^2x[/tex]

eventually giving me
[tex]\int tan^4xsec^3x dx[/tex] which i can't do

and if I used [tex]u^2=x+1[/tex] I eventually get

[tex]2\int u^3(u^2-1)^\frac{3}{2} du[/tex] which i also can't do

is there any useful substitution i can do?
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rock.freak667
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#2
Sep3-07, 07:28 PM
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ok well I found this formula
http://en.wikipedia.org/wiki/List_of...onal_functions (last one on the page) but now I have to find a way to derive it...any help on its derivation is kindly accepted
bob1182006
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#3
Sep3-07, 07:54 PM
P: 492
hm..it seems they did integration by parts with u=x^n and dv=square root but I can't follow it from there...

But why can't you do the first integral? you can just change tan, sec to sin and cos's and then do a substitution which will give you like..3 fractions of u to some power divided by u to some greater power.

rock.freak667
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#4
Sep3-07, 08:15 PM
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How to integrate a function with a square root in it


Quote Quote by bob1182006 View Post
hm..it seems they did integration by parts with u=x^n and dv=square root but I can't follow it from there...

But why can't you do the first integral? you can just change tan, sec to sin and cos's and then do a substitution which will give you like..3 fractions of u to some power divided by u to some greater power.
because then i will get

[tex]\int (sec^7\theta -2sec^5\theta + sec^3\theta)d\theta[/tex]
bob1182006
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#5
Sep3-07, 08:20 PM
P: 492
No just keep it as:

[tex]\int\frac{tan^4 x}{sec^3 x} dx[/tex]

but write tan and sec as sin/cos to give you:

[tex]\int\frac{sin^4 x}{cos^4 x}\frac{1}{cos^3} dx [/tex]

and then simplify those fractions into 1, then do a substitution.
rock.freak667
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#6
Sep3-07, 08:25 PM
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Quote Quote by bob1182006 View Post
[tex]\int\frac{sin^4 x}{cos^4 x}\frac{1}{cos^3} dx [/tex]

and then simplify those fractions into 1, then do a substitution.
Wouldn't simplifying it bring it back to an expression with tan and sec in it
bob1182006
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#7
Sep3-07, 08:28 PM
P: 492
no what do you get if you simplify [itex]cos^4 x * cos^3 x[/itex]? then you can do the substitution.
rock.freak667
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#8
Sep3-07, 08:32 PM
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Quote Quote by bob1182006 View Post
no what do you get if you simplify [itex]cos^4 x * cos^3 x[/itex]? then you can do the substitution.
Well I would get [tex] cos^3\theta-2sin^2\thetacos^2\theta+sin^4\thetacos^3\theta[/tex] by using [tex] cos^2\theta+sin^2\theta=1[/tex]

or should i get a substitution for [tex]cos^3 x[/tex]
bob1182006
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#9
Sep3-07, 08:34 PM
P: 492
if you have [itex]cos^n x * cos^m x[/itex] you simplify it the same as: [itex]u^n*u^m[/itex] which is equal to?
rock.freak667
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#10
Sep3-07, 08:38 PM
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[tex]u^n*u^m = u^{n+m}[/tex] thus [tex]cos^3\theta * cos^4\theta = cos^7\theta[/tex]
bob1182006
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#11
Sep3-07, 08:40 PM
P: 492
yep so now you have
[tex]\int\frac{sin^4 x dx}{cos^7 x}[/tex]

so what substitution would be nice there?
rock.freak667
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#12
Sep3-07, 08:44 PM
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Quote Quote by bob1182006 View Post
yep so now you have
[tex]\int\frac{sin^4 x dx}{cos^7 x}[/tex]

so what substitution would be nice there?
well normally I'd say [tex]u=cosx[/tex] but the high powers of the integrand is confusing me a bit
bob1182006
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#13
Sep3-07, 08:50 PM
P: 492
Hm..ok just saw a bit of a problem...

when you do u=cos x the top sin^4 x has to be split to sin x * sin^2 x * sin x which will be a square root * a polynomial.

also that reduction formula you found might not work since it needs x^(n-1) and if you use it you'll get x^(1/2) and x^(-1/2) so you wont hit x^0 which is needed....
rock.freak667
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#14
Sep3-07, 08:53 PM
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so...there really is no way to integrate this?
Mindscrape
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#15
Sep3-07, 09:01 PM
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There is an analytical answer, but it is a pretty nasty integral judging by the answer mathematica gives. I can't immediately think of a good trig substitution or good way to go about it. It might end up being that you need to use a mixture of integration by parts and some form of substitution, but if this is a problem that you set up I would make sure that everything leading up to the equation is right.
bob1182006
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#16
Sep3-07, 09:02 PM
P: 492
the integrator from the Mathematica website can do it but I have no idea how...

seems like you're going to get some sort of square root * polynomial and then + some hyperbolic function...
rock.freak667
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#17
Sep3-07, 09:04 PM
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Well this is just one question that my Further math teacher gave me..although he doesn't show you how to do them if you don't get it out
PowerIso
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#18
Sep3-07, 10:26 PM
P: 329
I got a solution, but damn it was hard. I am going to try to find an easier method.


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