Trigonometric Substitution

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mill4864

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Sep 17, 2009
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Evaluate the Integral of:

( dt ) / ( sqroot ( t^2 - 6t + 13 ) ).



I believe I have done all of the trigonometric substitution correctly (could be wrong, however). I think my problem is simplifying the final answer.



My work:


( dt ) / ( sqroot ( ( t - 3 )^2 + 4 )

u = t - 3
du = dt

( du ) / ( sqroot ( u^2 + 4 ) )

u = 2 tanx
du = 2 (secx)^2 dx
x = tan^-1 ( ( t - 3 ) / 2 ) < not sure about this ???

( 2 (secx)^2 ) / ( sqroot ( ( 2 tanx )^2) + 4 ) dx

Simplified into:

Integral of secx dx

= ln | secx + tanx | + c

= ln | sec ( tan^-1 ( ( t - 3 ) / 2 ) ) + tan ( tan^-1 ( ( t - 3 ) / 2 ) ) | + c

And that is where I don't know how to simplify. I could've made a mistake previously in the problem, also.
 
Hey, mill4864, you posted the identical problem on Sat, Sept. 19 which I solved for you. Are you that forgetful or do you suffer from some sort of dementia?
 
mill4864, see you previous thread and if there is something you don't understand about my analysis, then elucidate and I will attempt to set you straight. Putting up the same problem again is a waste of time and a waste of the board's memory.

Post Script: What's with the arc tangent crap?
 
It was a mistake. I apologize for reposting the same problem. After a posted the problem for the first time, I stopped doing my math homework and forgot that I had even posted it. When I started back up on my homework today, I was still on the same problem and was stuck and forgot about the post, hence me posting again. If it's really that big of a deal, just delete this thread.

And about the "arc tangent crap". After I had integrated, I had to put the solution in terms of t, not x. So I solved for x in terms of u, then I solved for u in terms of t. I didn't think about the fact that I could just solve for tan x.

This board is where people come to get help on math homework. If we already knew how to do the problem, we wouldn't come here for help. So taunting people for doing a problem wrong is not really appropriate and is very immature. But I suppose if it makes you feel better about yourself, trying to criticize people on the internet, then you really need to grow up and get a life.
 
Hey mill, I did the analysis on this problem and the check for you in your pervious thread.

Obviously you have a problem, as we are not here to babysit you.
 
Your work looks good.

\(\displaystyle \int sec(u)du=ln|sec(u)+tan(u)|\)

\(\displaystyle ln|sec(t-3)+tan(t-3)|+C\)

Now, you are done. Just leave it like that.
 
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