Here's the question you clicked on:
eistein+newton+gallileo=_me
find the integral of 1/root(-x^2 + 6x - 8) with explanations
thats the integral?
i got the information from here: :/ idk tho http://mathforum.org/library/drmath/view/64571.html
\[\int\frac{1}{\sqrt{-x^2+6x-8}}dx=\int\frac{1}{\sqrt{-x^2+6x-9+1}}dx=\int\frac{1}{\sqrt{1-(x-3)^2}}dx=\] \[=\arcsin{(x-3)}+C\]
no the answer is the arcsin one by nikvist
Use completing the square \[\rightarrow \int\limits_{}^{} \frac{dx}{\sqrt{1-(x-3)^{2}}}\] ...yeah thats what i got
CAN YOU PLEASE EXPLAIN
\[\int\frac{1}{\sqrt{1-x^2}}dx=\arcsin{x}+C\] basic integral
u = x-3 du = dx \[\rightarrow \int\limits_{}^{} \frac{du}{\sqrt{1-u^{2}}}\] u = sin(theta) du = cos(theta) \[\rightarrow \int\limits_{}^{}\frac{\cos(\theta)}{\sqrt{1-\sin^{2}(\theta)}}d \theta = \int\limits_{}^{}\frac{\cos(\theta)}{\sqrt{\cos^{2}(\theta)}}d \theta = \int\limits_{}^{}d \theta = \theta+C\] \[\theta = \sin^{-1}(u) = \sin^{-1}(x-3)\]
so automatically once i see such a question i should use trig substitutions ? what about this one 4/(4-x^2)
integral for arc sin is \[\int\limits_{}^{} du/\sqrt{a^2-u^2}\]