Hi! Thank you very much!

Question

My question is:

how do you find the antiderivative of 1 / (square root[1-(x^2)])

aka

antideriv( [1/(x^2)] ^ -1)

THank you!!

Answer 1

aka?

think of a right triangle, hypotenuse 1, opposite side x, adjacent side sqrt(1-x^2)

so cosineTheta=sqrt(1-x^2)
takeing derivative
- sinTheta dTheta/dx=1/sqrt(1-x^2) * -x

but x= sinTheta
-sinTheta dTheta=-sinTheta *dx/sqrt(1-x^2)

or dTheta= dx/sqrt( )
so the integral is now INT dTheta which is Theta. But Theta is arcsin x

so the answer is arcsinx