Let's assume I have this simple integral:
$\int \frac{-6}{\sqrt{1-x^2}} dx = -6 \int \frac{dx}{\sqrt{1-x^2}} = -6\arcsin{x} + C$
Question is: can I also write down the result as: $6\arccos{x}+C?$
And thus would the following be true?
$\int \frac{a}{\sqrt{b^2-x^2}}dx = a\arcsin{\frac{x}{b}} +C = -a\arccos{\frac{x}{b}}+C$
Best Answer
Yes (to the first question), since $(-\arcsin)'=\arccos'$. And the answer to the second question is also affirmative.