- #1

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## Homework Statement

How to you integrate (x

^{2}-5)

^{1/2}

## Homework Equations

## The Attempt at a Solution

I have tried substitution, integration by part, and none seems to work. I really don't know where to start.

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- Thread starter gaobo9109
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- #1

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How to you integrate (x

I have tried substitution, integration by part, and none seems to work. I really don't know where to start.

- #2

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make a substitution:

[tex] x=\sqrt{5} \cosh t [/tex]

[tex] x=\sqrt{5} \cosh t [/tex]

- #3

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- #4

uart

Science Advisor

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Well the cosh substitution is the easiest way, but you can also attack it as follows.

Rearrange the integrand as :

[tex] I = \int \frac{x^2-5}{\sqrt{x^2-5}}\, dx[/tex]

Split into two parts :

[tex] I = \int \frac{x^2}{\sqrt{x^2-5}} \, dx - \int \frac{5}{\sqrt{x^2-5}} \, dx[/tex]

You should be able to evaluate the second part by use of "standard integrals" and if you carefully apply integration by parts to the first integral you can reduce it to an expression minus "I" (where "I" is the original integral).

This lets you get it into the form of : "

I know that's only an outline, but it does work! See how far you can get with it.

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- #5

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[tex] I= \int \sqrt{x^2 -5} \, dx = \int \sqrt{x^2 +(i\sqrt{5})^2} \, dx \, , \, \forall x>\sqrt{5} [/tex]

Now you can make the substitution

[tex] x= i\sqrt{5} \tan t [/tex]

However, the road is pretty tough. The hyperbolic function shortens a lot of calculations.

- #6

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Or you could try u = √(5)sec*x*

- #7

uart

Science Advisor

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Or you could try u = √(5)secx

Would you care to elaborate on how that works Bohrok?

- #8

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use x = root5 sectheta...

i've solved it....works like a dream!....

i've solved it....works like a dream!....

- #9

uart

Science Advisor

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use x = root5 sectheta...

i've solved it....works like a dream!....

Ok so that substitution leads to :

[tex] 5 \, \int \frac{\sin^2(\theta)}{\cos^3(\theta)} \, d \theta[/tex]

Where did you go from there?

- #10

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yeah now

5[tex]

\int tan\theta tan\theta sec\theta d\theta

[/tex]

put t = [tex] sec\theta [/tex]

therefore dt = [tex] sec\theta tan\theta d\theta [/tex]

draw a right angled triangle.... using t = [tex] sec\theta[/tex].. you'll get value of two sides... find the value of the third side by pythagoras theorem... that way now you can find [tex] tan\theta [/tex]

put the value of [tex] tan\theta [/tex] and dt in the integrand..... solve !

5[tex]

\int tan\theta tan\theta sec\theta d\theta

[/tex]

put t = [tex] sec\theta [/tex]

therefore dt = [tex] sec\theta tan\theta d\theta [/tex]

draw a right angled triangle.... using t = [tex] sec\theta[/tex].. you'll get value of two sides... find the value of the third side by pythagoras theorem... that way now you can find [tex] tan\theta [/tex]

put the value of [tex] tan\theta [/tex] and dt in the integrand..... solve !

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- #11

uart

Science Advisor

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yeah now

5[tex]

\int tan\theta tan\theta sec\theta d\theta

[/tex]

put t = [tex] sec\theta [/tex]

therefore dt = [tex] sec\theta tan\theta d\theta [/tex]

draw a right angled triangle.... using t = [tex] sec\theta[/tex].. you'll get value of two sides... find the value of the third side by pythagoras theorem... that way now you can find [tex] tan\theta [/tex]

put the value of [tex] tan\theta [/tex] and dt in the integrand..... solve !

Is that meant to be a joke? It obviously leads you around in a big circle back to the original integral.

Look there were two perfectly good methods presented to solve this in replies #2 and #4. Enough of the junk replies please.

- #12

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could you please check it for me...

PS: moderators... i m not posting the solution... i am actually checking if what i've done is correct...

http://img831.imageshack.us/i/p1010910lr.jpg/

- #13

uart

Science Advisor

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could you please check it for me...

PS: moderators... i m not posting the solution... i am actually checking if what i've done is correct...

http://img831.imageshack.us/i/p1010910lr.jpg/

Hi rahuld.exe. Yes that still has an error, in the triangle you forgot the sqrt sign.

Where you got [itex]\sec(\theta) = t^2 + 1[/itex] it should have been [itex]\sec(\theta) = \sqrt{(t^2 + 1)}[/itex]. So you still end up with a difficult integral.

PS. Sorry if I was rude with the previous reply but I wasn't sure whether or not your effort was serious. Now I see you are making a genuine effort, the difficulty is just that the original [itex]x=\sqrt{5} \sec(\theta)[/tex] substitution isn't all that useful and that's what's making it hard for you to proceed with that method.

Take a look at the method I recommended in reply #4, you'll get it out in a just a few lines! In that method use the following "standard integral",

[tex]\int \frac{1}{\sqrt{x^2-a^2}} \, dx = \log_e(x + \sqrt{x^2 - a^2 })[/tex]

BTW The LHS above is actually inverse_cosh (hence the connection to bigubau's method). But most standard integral tables give it in the logarithmic form above, so we can keep completely clear of the hyperbolics with this method.

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- #14

uart

Science Advisor

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[tex] \int \frac{x^2}{\sqrt{x^2-5}} \, dx = \int x \, \frac{d}{dx}\left\{\sqrt{x^2-5}\right\} \, dx [/tex]

- #15

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