Trying to do an anti-derivative

[PowerMacG5]

My physics teacher gave us an integration to examine, so we could use it to find the electric field of a line of charge. The anti-derivative he wants us to find is this. He already told us the answer is this.

I already tried applying everything I know, and can't get anywhere. Whether it's we didn't learn it in my calculus class yet, or I just can't figure it out remains to be seen. I tried doing a u substitution, which comes to a dead end. So then I tried integration by parts. I tried to let u and dv equal various things, but it just seems to make it more and more complicated. Just for your information, the last topic we did was integration by parts in my calculus class, so if you think the technique to solve it is past what we did, could you still post it so I can examine it? Thanks in advance.

 

[PowerMacG5]

Bump.

 

[goldsystems]

Well, you're answer is right... Maple knows all. How to work it out, well ..heh, that involves getting up, sorry man.

 

[RossGr]

This works well with a trig substitution of y=atanx. Have you done trig substitutions yet?

 

[PowerMacG5]

Originally posted by: RossGr
This works well with a trig substitution of y=atanx. Have you done trig substitutions yet?

Nope, that could be why I couldn't solve it. Would you mind posting the steps, or do you think my text would do a better job explaining it?

 

[RossGr]

let y = a tanx

dy = a sec^2 (x) dx

So the denominator of the integral becomes (a^2 + (a Tan x)^2)^3/2

factor to get a^3 (1 + tan^2 x) = a^3 sec^3 x

replaced dy with dx results in integral ( 1/sec x) dx = Int (cos x dx) = sin x + C

(don't forget that a^2 in the denominator! I didn't carry it through just because it is messy enough!)

So we have as an answer of (Sin x)/a^2 + C

We now need to undo the substitution

y= a Tan x => y/a = Tan x


edit: catching some typos!


Remember the basic definition of Tan x = (opposite / adjacent) This implies y= opposite side, a = adjacent side so the hypotenuse must be Sqrt ( a^2 + y^2) This means

Sin x = opposite/ hypot. = y/sqrt(a^2 +y^2)

final answer

y/(a^2 (sqrt (a^2 + y^2))

 

I hate calculus.

Okay, continue.

 

[PowerMacG5]

Originally posted by: RossGr
let y = a tanx

dy = a sec^2 (x) dx

So the denominator of the integral becomes (a^2 + (a Tan x)^2)^3/2

factor to get a^3 (1 + tan^2 x) = a^3 sec^3 x

replaced dy with dx results in integral ( 1/sec x) dx = Int (cos x dx) = sin x + C

(don't forget that a^2 in the denominator! I didn't carry it through just because it is messy enough!)

So we have as an answer of (Sin x)/a^2 + C

We now need to undo the substitution

y= a Tan x => y/a = Tan x


edit: catching some typos!


Remember the basic definition of Tan x = (opposite / adjacent) This implies y= opposite side, a = adjacent side so the hypotenuse must be Sqrt ( a^2 + y^2) This means

Sin x = opposite/ hypot. = y/sqrt(a^2 +y^2)

final answer

y/(a^2 (sqrt (a^2 + y^2))

Wow, thank you so much.