Integ 1 / sqr root(a^2 - x^2 )

[teng125]

does anybody knows how to do integ 1 / sqr root(a^2 - x^2 )...
pls help...

 

[Mindscrape]

Think inverse trig.

 

[teng125]

ya i thought of using diff of arcsin ax = a / sqr root (1 - (ax)^2 )
but if i do so the numerator will be 1/a^2 rite??

pls help

 

[Mindscrape]

Close, it would be 1/a, like this: $$\int\frac{du}{\sqrt{a^2-u^2}} = arcsin \frac{u}{a} + C$$