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

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The discussion focuses on solving the integral of 1 over the square root of (a^2 - x^2). Participants suggest using inverse trigonometric functions, particularly arcsin, to approach the problem. One user clarifies that the correct form of the integral leads to arcsin(u/a) + C, rather than the initially considered 1/a^2. The conversation emphasizes the importance of proper substitution in integration. Overall, the integral can be effectively solved using the arcsin function.
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does anybody knows how to do integ 1 / sqr root(a^2 - x^2 )...
pls help...
 
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Think inverse trig.
 
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
 
Close, it would be 1/a, like this: \int\frac{du}{\sqrt{a^2-u^2}} = arcsin \frac{u}{a} + C
 
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