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Another integrationby frozen7
Tags: integration 
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#1
Dec2805, 08:10 AM

P: 167

[tex]\int \frac{1}{x^2 +4} [/tex] How to integrate this without knowing the derivatives of inverse function?



#2
Dec2805, 08:14 AM

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P: 12,016

If you don't know about an antiderivative of the integrand, or are unable to transform your integrand in such a manner that an antiderivative becomes apparent, then the fundamental theorem of calculus is of minor use to you in that particular case in the evaluation of your integral.



#3
Dec2805, 08:46 AM

P: 91

I think he/she is asking whether you need to know inverse trig derivatives/integrals to do this problem. You don't need to know the derivative of inverse trigonometric functions to do this problem, however you need to know about trig substitutions, if you can do trig substitutions then this integral can be done in two steps, one is a regular substitution that gets it in a form such as 1/(u^2 +1) and another trig substitution after that.



#4
Dec2805, 08:56 AM

P: 390

Another integration
If you facter (x^2 + 4) you get (x + 2i)(x  2i) where i is the complex number, sqrt(1). From there you can decompose 1/(x^2 + 4) into two fractions: 1/(x^2 + 4) = A/(x + 2i) + B/(x  2i) And you'll end up with a complex logarithm as the answer which will fit the definition of arctan, if I'm not mistaken. (Or something like that) Either way plugging numbers in will still get you the same results as the real answer. :) 


#5
Dec2805, 09:00 AM

P: 167




#6
Dec2805, 09:06 AM

P: 91

I'll assume you can do the first step, then use the trig substitution x = tan(θ). Remember after you have your answer you have to get it back in terms of x, your answer will be in terms of theta.
Edit: forgot a comma and a then, it might make it confusing. 


#7
Dec2805, 09:34 AM

HW Helper
P: 1,421

@frozen7: You should look it up in your textbook or try here So, what's the substitution? Is it [itex]x = \tan \theta[/itex]? Or what is it? 


#8
Dec2805, 10:03 AM

P: 223

in the integral, you're missing the dx.
draw a triangle, it should help you see which values to substitute. 


#9
Dec2805, 11:59 AM

P: 80

At first [tex]\int \frac 1 {x^2+1} dx = \arctan(x)[/tex] Now put [tex]u=2x[/tex] and you'll get [tex] \frac 1 8 \int \frac 1 {u^2+1} du = 1/8 \arctan(2x)[/tex] 


#10
Dec2805, 12:16 PM

P: 4,006

maverick, the substitution is not u=2x but u=x/2
[tex]du = \frac{dx}{2}[/tex] [tex]\frac{1}{x^2+4} = \frac{1}{4(\frac{x^2}{4} + 1)} = \frac{1}{4( ( \frac{x}{2} )^2 + 1)}[/tex] marlon 


#11
Dec2805, 12:22 PM

P: 80

Then correction: [tex] \frac 1 2 \int \frac 1 {u^2+1} du = \arctan(x/2)[/tex] hm, this looks smarter. 


#12
Dec2805, 12:24 PM

P: 4,006

marlon 


#13
Dec2805, 12:29 PM

P: 80

[tex]1/2 \arctan(x/2)[/tex] maybe i need sleep. It's 3:30am. 


#14
Dec2805, 12:34 PM

P: 4,006

Here in Western Europe it's 19.30 pm and i am looking foreward to attend the SAW2 premiere tonight. OOHH YES, THERE WILL BE BLOOD Sleep well :) marlon 


#15
Dec2805, 04:46 PM

P: 80




#16
Dec2905, 01:45 AM

P: 390

What's up with all of the arctans? Why not be frank with the function and do what you gotta do..... use partial fractions!?



#17
Dec2905, 02:46 AM

P: 1,076

[tex]\int \frac 1 {u^2+a^2} du = \frac 1 {a} \arctan( \frac u {a} ) + C[/tex] 


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