Question

find the integral of 1/root(-x^2 + 6x - 8)
with explanations

Answer 1

thats the integral?

Answer 2

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

Answer 3

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

Answer 4

no the answer is the arcsin one by nikvist

Answer 5

Use completing the square
\[\rightarrow \int\limits_{}^{} \frac{dx}{\sqrt{1-(x-3)^{2}}}\]

...yeah thats what i got

Answer 6

CAN YOU PLEASE EXPLAIN

Answer 7

he wants you to explain

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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