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

by rock.freak667
Tags: function, integrate, root, square
 HW Helper P: 6,214 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?
 HW Helper P: 6,214 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
 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.
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P: 6,214

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

 Quote by 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.
because then i will get

$$\int (sec^7\theta -2sec^5\theta + sec^3\theta)d\theta$$
 P: 492 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.
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P: 6,214
 Quote by bob1182006 $$\int\frac{sin^4 x}{cos^4 x}\frac{1}{cos^3} dx$$ 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
 P: 492 no what do you get if you simplify $cos^4 x * cos^3 x$? then you can do the substitution.
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P: 6,214
 Quote by bob1182006 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 $$cos^3 x$$
 P: 492 if you have $cos^n x * cos^m x$ you simplify it the same as: $u^n*u^m$ which is equal to?
 HW Helper P: 6,214 $$u^n*u^m = u^{n+m}$$ thus $$cos^3\theta * cos^4\theta = cos^7\theta$$
 P: 492 yep so now you have $$\int\frac{sin^4 x dx}{cos^7 x}$$ so what substitution would be nice there?
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P: 6,214
 Quote by bob1182006 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
 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....
 HW Helper P: 6,214 so...there really is no way to integrate this?
 P: 1,877 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.
 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...
 HW Helper P: 6,214 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
 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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