What's $\int \frac{1}{\sqrt{25-x^2}}$

Question

What is $$\int \frac{1}{\sqrt{25-x^2}}$$ WolframAlpha says $\sin^{-1}(\frac{x}{5})$ while I got $\frac{1}{5}\sin^{-1}(\frac{x}{5})$. What is correct? Thanks in advance.

Answer 1

Let $x = 5u \to dx = 5du \to \displaystyle \int \dfrac{dx}{\sqrt{25-x^2}} = \displaystyle \int \dfrac{5du}{\sqrt{25-25u^2}} = \displaystyle \int \dfrac{du}{\sqrt{1-u^2}} =\arcsin u + C = \arcsin \left(\frac{x}{5}\right) + C$

Answer 2

Hint:-

Substitute $x=5\sin\theta$ or $5\cos \theta$.