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How to integrate a function with a square root in it 
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#1
Sep307, 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^21)^\frac{3}{2} du[/tex] which i also can't do is there any useful substitution i can do? 


#2
Sep307, 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 


#3
Sep307, 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. 


#4
Sep307, 08:15 PM

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How to integrate a function with a square root in it
[tex]\int (sec^7\theta 2sec^5\theta + sec^3\theta)d\theta[/tex] 


#5
Sep307, 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. 


#6
Sep307, 08:25 PM

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#7
Sep307, 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.



#8
Sep307, 08:32 PM

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or should i get a substitution for [tex]cos^3 x[/tex] 


#9
Sep307, 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?



#10
Sep307, 08:38 PM

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P: 6,202

[tex]u^n*u^m = u^{n+m}[/tex] thus [tex]cos^3\theta * cos^4\theta = cos^7\theta[/tex]



#11
Sep307, 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? 


#12
Sep307, 08:44 PM

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#13
Sep307, 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^(n1) and if you use it you'll get x^(1/2) and x^(1/2) so you wont hit x^0 which is needed.... 


#14
Sep307, 08:53 PM

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so...there really is no way to integrate this?



#15
Sep307, 09:01 PM

P: 1,874

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.



#16
Sep307, 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... 


#17
Sep307, 09:04 PM

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P: 6,202

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



#18
Sep307, 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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