Question

\[\int \frac{dt}{\sqrt(t^2 -6t +13)}\]

Answer 1

is the idea to complete the square the radical?

Answer 2

i was thinking trig substitution...but don't know how...(that's the section)

Answer 3

you have one step before the trig sub
write \(x^2-6x+13=(x-3)^2+4\) to get
\[\int\frac{dt}{\sqrt{(t-3)^2+4}}\] and the call say \(u=t-3\) and finally a trig sub

Answer 4

at this point use usual one for \(\sqrt{u^2+b^2}\) use \(u=b\tan(\theta)\)

Answer 5

\[u=bsec(\theta)\] ?

Answer 6

maybe it does not matter
lets check

Answer 7

\[1+tan^2(\theta)=sec^2(\theta)\]
\[\sqrt(1+tan^2(\theta))=sec(\theta)\]
b=2

Answer 8

\[\sqrt{u^2+4}\] \(u=2\tan(\theta), du=\sec^2(\theta)d\theta\)
\[\sqrt{(2\tan^2(\theta)+4}=\sqrt{4(tan^2(\theta)+1)}=2\sqrt{sec^2(\theta)}=2\sec(\theta)\]
and the integral becomes
\[\int\frac{\sec^2(\theta)}{\sec(\theta)}d\theta=\int\sec(\theta)d\theta\]

Answer 9

oh doh not i am not, the two cancels nvm

Answer 10

thanks satellite

Answer 11

\[\int \frac{1}{\sqrt{t^2-6t+13}}dt\]\[=\int \frac{1}{\sqrt{t^2-6t+3^3-3^2+13}}dt\]\[=\int \frac{1}{\sqrt{(t-3)^2+4}}dt\]
Let \(t-3 = 2tan\theta\)
\(dt = 2sec^2\theta d\theta\)
\[=\int \frac{1}{\sqrt{(2tan\theta)^2+4}}(2sec^2\theta d\theta)\]\[=\int \frac{2sec^2\theta}{\sqrt{(2tan\theta)^2+4}} d\theta\]\[=\int \frac{2sec^2\theta}{\sqrt{(4(tan^2\theta+1)^2}} d\theta\]\[=\int \frac{2sec^2\theta}{\sqrt{(2sec^2\theta)^2}} d\theta\]\[=\int \frac{2sec^2\theta}{2sec\theta} d\theta\]\[=\int sec\theta d\theta\]Amazing :D

Answer 12

Thanks @Callisto :)