## eistein+newton+gallileo=_me 4 years ago find the integral of 1/root(-x^2 + 6x - 8) with explanations

1. LagrangeSon678

thats the integral?

2. Izzy25

i got the information from here: :/ idk tho http://mathforum.org/library/drmath/view/64571.html

3. nikvist

$\int\frac{1}{\sqrt{-x^2+6x-8}}dx=\int\frac{1}{\sqrt{-x^2+6x-9+1}}dx=\int\frac{1}{\sqrt{1-(x-3)^2}}dx=$ $=\arcsin{(x-3)}+C$

4. eistein+newton+gallileo=_me

no the answer is the arcsin one by nikvist

5. dumbcow

Use completing the square $\rightarrow \int\limits_{}^{} \frac{dx}{\sqrt{1-(x-3)^{2}}}$ ...yeah thats what i got

6. eistein+newton+gallileo=_me

7. Izzy25

he wants you to explain

8. nikvist

$\int\frac{1}{\sqrt{1-x^2}}dx=\arcsin{x}+C$ basic integral

9. dumbcow

u = x-3 du = dx $\rightarrow \int\limits_{}^{} \frac{du}{\sqrt{1-u^{2}}}$ u = sin(theta) du = cos(theta) $\rightarrow \int\limits_{}^{}\frac{\cos(\theta)}{\sqrt{1-\sin^{2}(\theta)}}d \theta = \int\limits_{}^{}\frac{\cos(\theta)}{\sqrt{\cos^{2}(\theta)}}d \theta = \int\limits_{}^{}d \theta = \theta+C$ $\theta = \sin^{-1}(u) = \sin^{-1}(x-3)$

10. eistein+newton+gallileo=_me

so automatically once i see such a question i should use trig substitutions ? what about this one 4/(4-x^2)

11. LagrangeSon678

integral for arc sin is $\int\limits_{}^{} du/\sqrt{a^2-u^2}$