I complete the square for let (x-3) = 2 sec theta and dx = 2 sec theta tan theta dtheta

is this correct? wolfram did it differently so I just want to check to see if my way is correct as well.

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- March 4th 2014, 12:09 PMJonroberts74trig sub correct or incorrect
I complete the square for let (x-3) = 2 sec theta and dx = 2 sec theta tan theta dtheta

is this correct? wolfram did it differently so I just want to check to see if my way is correct as well. - March 4th 2014, 12:36 PMromsekRe: trig sub correct or incorrect
your way looks correct as well. The plots of your function and wolfram's integral lie on top of each other.

I have to say though that this is clearly a partial fractions problem and not a trig substitution problem even if you can find an substitution that works.

$x^2-6x+5=(x-5)(x-1)$ so a bit of work shows

$\dfrac{1}{x^2-6x+5}=\dfrac{1}{4}\left(\dfrac{1}{x-5}-\dfrac{1}{x-1}\right)$

and the integral pops right out. - March 4th 2014, 12:50 PMJonroberts74Re: trig sub correct or incorrect
ah that makes much more sense. its from a improper integral convergence or divergence question. thanks