# How to integrate a function with a square root in it

1. Sep 3, 2007

### rock.freak667

1. The problem statement, all variables and given/known data

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

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 $$u=tan^2x$$ 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?

2. Sep 3, 2007

3. Sep 3, 2007

### 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.

Last edited: Sep 3, 2007
4. Sep 3, 2007

### rock.freak667

because then i will get

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

5. Sep 3, 2007

### 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.

6. Sep 3, 2007

### rock.freak667

Wouldn't simplifying it bring it back to an expression with tan and sec in it

7. Sep 3, 2007

### bob1182006

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

8. Sep 3, 2007

### rock.freak667

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 $$cos^3 x$$

9. Sep 3, 2007

### bob1182006

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

10. Sep 3, 2007

### rock.freak667

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

11. Sep 3, 2007

### bob1182006

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

so what substitution would be nice there?

12. Sep 3, 2007

### rock.freak667

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

13. Sep 3, 2007

### 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 wont hit x^0 which is needed....

14. Sep 3, 2007

### rock.freak667

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

15. Sep 3, 2007

### 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.

16. Sep 3, 2007

### 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...

17. Sep 3, 2007

### 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

18. Sep 3, 2007

### PowerIso

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

19. Sep 4, 2007

### 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.

20. Sep 4, 2007

### 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).

Last edited: Sep 4, 2007