I am paying fg for this question to be done out right and all of the steps shown, I will pay generously or an agreed upon payment either way.

you have to prove this using TRIG Substitution

(dx)/(x^2+a^2) = (1/a)tan^-1(x/a)+c

yet again post or pm me and all steps must be shown and labeled what you did at each step

thanks

edit:

also have other questions that I would be willing to pay fg for help with them, as long as this question goes well.

This post was edited by Acdc-rocks[tom] on Jul 27 2013 10:26am

Where to start from ?

Do you know the rule :

(f^-1)' = 1 / ( f'(f^-1)) ?

If so :

(tan^-1)' = 1 / (tan'(tan^-1))

with tan' = (sin/cos)' = (cos² + sin²) / cos² = 1 + tan²

hence (tan^-1)' = 1 / ( 1 + tan²(tan^-1)) = 1 / ( 1 + x² )

Now compose with x-> x/a :

(tan^-1 (x/a)) = (1/a) * 1 / ( 1 + (x/a)² ) = a / ( x² + a² )

Hence dx / (x² +a²) = (1/a). tan^-1(x/a) + C

I can develop any step, just let me know what is still unclear.

Good luck !

thanks for the help, although this isnt in the trig sub, it did help a little bit. sending you some gold for your efforts.

still need help getting this done in trig substitution.

Thanks for the donation.

Maybe this way ?

First, establish that (tan x) ' = 1 + tan² (x)

For x stricly between -pi/2 and pi/2, we must find the limit when h approaches zero of :

Dh = (tan(x+h) - tan(x)) / h

Dh = ( sin(x+h)/cos(x+h) - sin x / cos x ) / h

Dh = ( sin (x+h).cos x - sin x.cos(x+h) ) / ( h.cos x.cos (x+h) )

and you can develop :
sin (x+h) = sin x.cos h + sin h. cos x
cos (x+h) = cos x.cos h - sin x.sin h

Plug this into the previous formula :

Dh =( ( sin x.cos h + sin h.cos x ).cos x - (cos x.cos h - sin x.sin h).sin x ) / ( h.cos x.cos (x+h) )

After cancellation :

Dh = ( cos² x + sin² x ). sin h / ( h.cos x.cos (x+h) )
Dh = ( sin h / h ).( 1 / (cos x.cos (x+h) )

The first parenthesis has limit 1 (easy to establish with a figure).
And the second has obviously limit 1 / cos² x,
and 1 / cos² x = 1 + tan² x

Conclusion : Dh has 1 + tan² x for limit, hence (tan x) ' = 1 + tan² x

Second step : let's work with tan^-1, with range the real numbers and target set ] -pi/2 ; pi/2 [
And please, allow me to use Arctan instead of tan^-1 (I'm not used to tan^-1, that can easily be confused with 1 / tan).

Let Delta_h be ( Arctan (x+h) - Arctan x ) / h, for a given x and a variable h.
We must find the limit (if it exists) of Delta_h when h approaches zero.

Let y = Arctan x and y + k = Arctan (x+h), so that tan y = x and tan (y+k) = x+h

Delta_h = ( (y+k) - y ) / h = k / h

And since we know the derivative function of tan, we can use a Taylor developement of order 1 :

tan (y+k) = tan y + (1+tan² y).k + O(k²)

so :
x + h = x + ( 1+x²).k + O(k²)
h = k.(1 + x² + O(k))

hence : k / h = 1 / ( 1 + x² + O(k) )

obviously k has limit zero when h approaches zero, so Delta_h has for limit 1 / ( 1+x²)

This proves that (Arctan x) ' = 1 / (1+x²)