Question

Find the indefinite integral:
Int dx/sqrt(8+2x-x^2)

Answer 1

\[\int\limits_{}^{}\frac{1}{\sqrt{8+2x-x^2}} dx\]

Answer 2

trig subs i believe

Answer 3

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

Answer 4

Then a u-sub on (x-1) then a trig sub

Answer 5

Or recognize its in the arcsin form:
\[\int\limits \frac{du}{\sqrt{a^2-u^2}}=\arcsin(\frac{u}{a})+C\]

Answer 6

then plug in values for u and a and ur done

Answer 7

let \[\cos(\theta)=\frac{x-1}{3}\]
=> \[\cos^2(\theta)=\frac{(x-1)^2}{9}=> 9\cos^2(\theta)=(x-1)^2\]
\[-\sin(\theta) d \theta=\frac{1}{3} dx\]
\[\int\limits_{}^{}\frac{-3\sin(\theta)}{\sqrt{9-9\cos^2(\theta)}}d \theta\]
\[-3\int\limits_{}^{}\frac{\sin(\theta)}{\sqrt{9}\sqrt{1-\cos^2(\theta)}} d \theta\]
\[-\int\limits_{}^{}\frac{\sin(\theta)}{\sin(\theta)} d \theta=-\int\limits_{}^{}1 d \theta=- \theta+C\]
remember \[\cos(\theta)=\frac{x-1}{3} =>\theta=\cos^{-1}(\frac{x-1}{3})\]

Answer 8

\[-\theta+C=-\cos^{-1}(\frac{x-1}{3})+C\]

Answer 9

wait, why did I get arcsin(x-1/3)+c?

Answer 10

arcsin((x-1)/3)+C or -arccos((x-1)/3)+C is fine