Integrating \frac{1}{x^2+4} - Solution

[frozen7]

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

 

[arildno]

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

 

[what]

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.

 

[Pseudo Statistic]

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

You can use some basic algebra and the method of partial fractions..
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. :)

 

[frozen7]

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.

Could you explain further?

 

[what]

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.

 

[VietDao29]

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.

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

 

[andrewchang]

in the integral, you're missing the dx.

draw a triangle, it should help you see which values to substitute.

 

[maverick6664]

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

If you know integral of inverse function it's easy.
At first \int \frac 1 {x^2+1} dx = \arctan(x)

Now put u=2x and you'll get

\frac 1 8 \int \frac 1 {u^2+1} du = 1/8 \arctan(2x)

 

[marlon]

maverick, the substitution is not u=2x but u=x/2

du = \frac{dx}{2}
\frac{1}{x^2+4} = \frac{1}{4(\frac{x^2}{4} + 1)} = \frac{1}{4( ( \frac{x}{2} )^2 + 1)}

marlon

 

[maverick6664]

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

oh my bad! Thank you! I'm a bit euphoric because my favorite female speed skater Tomomi Okazaki was chosen for Trino Olympic

Then correction:

\frac 1 2 \int \frac 1 {u^2+1} du = \arctan(x/2)

hm, this looks smarter.

 

[marlon]

oh my bad! Thank you!
Then correction:
\frac 1 2 \int \frac 1 {u^2+1} du = \arctan(x/2)

I don't want to be whinning but the right hand side of your equation is not correct either. You are forgetting something :)

marlon

 

[maverick6664]

I don't want to be whinning but the right hand side of your equation is not correct either. You are forgetting something :)
marlon

ooops! lol

1/2 \arctan(x/2)

maybe i need sleep. It's 3:30am.

 

[marlon]

maybe i need sleep. It's 3:30am.

Aha, that happens to me too.

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

 

[maverick6664]

Aha, that happens to me too.
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

Got up now..but it's still 7:40am..I'll sleep again.. New Years day my family will pay first visit to a nearby shrine and my father's grave, and from Jan.2 I'll be "Home alone" because my wife and daughters will be in my wife's home town for nearly a week. Still sleepy... nite... (looks like I'm hijacking this thread.. this is a users' forum...) zzzzz...

 

[Pseudo Statistic]

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

 

[d_leet]

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

Because
\int \frac 1 {u^2+a^2} du = \frac 1 {a} \arctan( \frac u {a} ) + C

 

[marlon]

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

Because than you would be needing complex functions. That's against the rules if the calculus must remain "real"...

\frac{1}{x^2+a^2} = \frac{1}{x+ia}\frac{1}{x-ia}

marlon

 

[what]

Here is the solution i was hinting at, without knowing the derivative of inverse tangent...
\int \frac{1}{x^2+4} dx \ u = x/2 \ du =1/2 dx <br />
1/2 \int \frac{1}{u^2+1} du \ u =\tan(\theta) \ du = \sec^2(\theta) d\theta <br />
= 1/2 \int \frac{\sec^2(\theta)}{\tan^2(\theta) +1} d\theta = 1/2 \int \frac{\sec^2(\theta)}{\sec^2(\theta)} d\theta = 1/2 \int 1 \ d\theta<br />
= 1/2(\theta + C)
Now we have to bring back in terms of x so we use the way we defined the substitutions to help us..
\theta = \arctan(u) = \arctan(x/2)
therefore...
\int \frac{1}{x^2+4} dx = 1/2\arctan(x/2) +C

 

[Pseudo Statistic]

Because than you would be needing complex functions. That's against the rules if the calculus must remain "real"...
\frac{1}{x^2+a^2} = \frac{1}{x+ia}\frac{1}{x-ia}
marlon

It will-- you could then convert back to arctan using the definition of the function from Oiler's formula... the final answer will always be real wherever arctan is defined as real.. so why not?

 

[d_leet]

It will-- you could then convert back to arctan using the definition of the function from Oiler's formula... the final answer will always be real wherever arctan is defined as real.. so why not?

Because I doubt the original poster would know how to do that.

 

[andrewchang]

Here is the solution i was hinting at, without knowing the derivative of inverse tangent...
\int \frac{1}{x^2+4} dx \ u = x/2 \ du =1/2 dx <br />
1/2 \int \frac{1}{u^2+1} du \ u =\tan(\theta) \ du = \sec^2(\theta) d\theta <br />
= 1/2 \int \frac{\sec^2(\theta)}{\tan^2(\theta) +1} d\theta = 1/2 \int \frac{\sec^2(\theta)}{\sec^2(\theta)} d\theta = 1/2 \int 1 \ d\theta<br />
= 1/2(\theta + C)
Now we have to bring back in terms of x so we use the way we defined the substitutions to help us..
\theta = \arctan(u) = \arctan(x/2)
therefore...
\int \frac{1}{x^2+4} dx = 1/2\arctan(x/2) +C

that's the answer that i got, too

 

[marlon]

It will-- you could then convert back to arctan using the definition of the function from Oiler's formula... the final answer will always be real wherever arctan is defined as real.. so why not?

How would you do that and still get rid of the complex variable ?
marlon