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Jbreezy
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Homework Statement
Integral of dt/ (√(t^2 -6t + 13)
Homework Equations
I sub
v = t-3, and v = 2tan(θ)
The Attempt at a Solution
first I completed the square of t^2 -6t + 13, I got (t-3)^2 +4
Also I say v = t-3
∫ dv/ (√(v^2 +4)
I then sub in trig
∫(2sec^2(θ)dθ / 2√(tan^2(θ) +1) = ∫ (sec^2(θ)/ secθ)dθ
= ln|secθ +tanθ| = ln|(√v^2+4)/2 + v/2| edit sec here
= ln| √((t-3)^2 +4/2) + (t-3)/2| +c
I feel it isn't correct. What did I do?
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