Integral of $\int\ \sqrt{1-x^2} dx $ with 2 method

Question

I was asked to solve the indefinite integral using two method

$$\int\ \sqrt{1-x^2} dx $$

if I let x=sin $\theta$,

$$\int \sqrt{1-x^2}\, dx = \frac12 x\sqrt{1-x^2} + \frac12 \sin^{-1}x + C$$ ...(1)

als0, if I let x=cos $\theta$,

$$\int \sqrt{1-x^2}\, dx = \frac12 x\sqrt{1-x^2} - \frac12 \cos^{-1}x + C$$ ...(2)

this should be the same no matter what I choose x to be but looks different as (1)-(2) != 0

Any help would be appreciated

Answer 1

But they're equal, because$$\bigl(\forall x\in[-1,1]\bigr):\arcsin(x)+\arccos(x)=\frac\pi2.$$Therefore, the difference between your answers is constant.