Question

FOR PEOPLE HAVING DIFFICULTIES WITH MEMORIZING THE INVERSE TRIGONOMETRIC FUNCTIONS IN INTEGRALS!

how to prove that \(\Large \int \frac{1}{\sqrt{a^2 - x^2}} dx = \sin^{-1} \frac{x}{a} ?\) check the comment to see (Just give me a few minutes i'm a slow typer)

Answer 1

put x = acosx

Answer 2

sorry.... x =a cost

Answer 3

\[x=asin\theta\]

Answer 4

first...draw a right triangle with all the variables present...note that it's the square root of a constant minus square of a variable...so the constant is our hypotenuse of our right triangle



since the equation is \(\sqrt {a^2 -x^2}\) we can say that one of the legs is the \(\sqrt x^2\) so



remember to always put the x away from the bottom leg...with this..we can say that the last leg is the denominator so