How to integrate a function with a square root in it

[rock.freak667]

Homework Statement

find $$\int x^{\frac{3}{2}}\sqrt{1+x} dx$$

Homework Equations

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 $$u=tan^2x$$ and $$du = 2tanxsec^2x$$

eventually giving me
$$\int tan^4xsec^3x dx$$ which i can't do

and if I used $$u^2=x+1$$ I eventually get

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

is there any useful substitution i can do?

 

[rock.freak667]

ok well I found this formula
http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_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]

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]

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

$$\int (sec^7\theta -2sec^5\theta + sec^3\theta)d\theta$$

 

[bob1182006]

No just keep it as:

$$\int\frac{tan^4 x}{sec^3 x} dx$$

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

$$\int\frac{sin^4 x}{cos^4 x}\frac{1}{cos^3} dx$$

and then simplify those fractions into 1, then do a substitution.

 

[rock.freak667]

$$\int\frac{sin^4 x}{cos^4 x}\frac{1}{cos^3} dx$$

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]

no what do you get if you simplify $cos^4 x * cos^3 x$? then you can do the substitution.

 

[rock.freak667]

no what do you get if you simplify $cos^4 x * cos^3 x$? then you can do the substitution.

Well I would get $$cos^3\theta-2sin^2\thetacos^2\theta+sin^4\thetacos^3\theta$$ by using $$cos^2\theta+sin^2\theta=1$$

or should i get a substitution for $$cos^3 x$$

 

[bob1182006]

if you have $cos^n x * cos^m x$ you simplify it the same as: $u^n*u^m$ which is equal to?

 

[rock.freak667]

$$u^n*u^m = u^{n+m}$$ thus $$cos^3\theta * cos^4\theta = cos^7\theta$$

 

[bob1182006]

yep so now you have
$$\int\frac{sin^4 x dx}{cos^7 x}$$

so what substitution would be nice there?

 

[rock.freak667]

yep so now you have
$$\int\frac{sin^4 x dx}{cos^7 x}$$

so what substitution would be nice there?

well normally I'd say $$u=cosx$$ but the high powers of the integrand is confusing me a bit

 

[bob1182006]

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 won't hit x^0 which is needed...

 

[rock.freak667]

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

 

[Mindscrape]

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]

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]

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]

I got a solution, but damn it was hard. I am going to try to find an easier method.

 

[SanjeevGupta]

Tried x=(sinh(u))^2 and it works through to the same result as Mathematica.
After substitution you convert the sinh's, cosh's to exponents using sinh(u)=0.5*(exp(u)-exp(-u)), cosh(u)=0.5*(exp(u)+exp(-u)).
You get a simple integral but then you have to work to convert the sinh's and cosh's of multiples of u to powers of sinh(u) and cosh(u) but it works out fine.

 

[P.O.L.A.R]

How did you get u=tan^2(x)??

because x is really [sqrt(x)]^2 having trouble seeng the tan^2(x).

 

[rock.freak667]

Tried x=(sinh(u))^2 and it works through to the same result as Mathematica.
After substitution you convert the sinh's, cosh's to exponents using sinh(u)=0.5*(exp(u)-exp(-u)), cosh(u)=0.5*(exp(u)+exp(-u)).
You get a simple integral but then you have to work to convert the sinh's and cosh's of multiples of u to powers of sinh(u) and cosh(u) but it works out fine.

Well I didn't really learn hyperbolic functions yet as I am a few classes behind so I guess I'll have to read about it online and see if i can do it

 

[rock.freak667]

Well using $$x=sinh^2x$$ gives a lot of algebra to work out so I guess I must work it out and post back the answer