substitution integration question I want to integrate $\int \sqrt{1 - x^2} dx $.
When I substitute $x = \sin θ$ , I get the right answer. ( $ \cos^2\theta$ before integration)
But when I substitute $x = \cos θ$ , I don't get the right answer. ( $ -\sin^2\theta$ before integration).
What step is wrong here? If I proceed like this, don't I end up with a wrong answer?

The short answer is they become the same answer once you back-substitute $\theta$ for $x$. 
It makes more sense if you actually do the integrals.
$$\begin{align} \int \cos^2 \theta \,d\theta &= \frac{1}{2}\int (1 +\cos 2\theta)\,d\theta \\&= \frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= \frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= \frac{1}{2}\arcsin x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}$$
$$\begin{align} \int -\sin^2 \theta \,d\theta &= -\frac{1}{2}\int (1 -\cos 2\theta)\,d\theta \\&= -\frac{1}{2}\theta + \frac{1}{4}\sin 2\theta + C \\&= -\frac{1}{2}\theta + \frac{1}{2}\sin\theta\cos\theta + C \\&= -\frac{1}{2}\arccos x + \frac{1}{2}x\sqrt{1-x^2} + C \end{align}$$
The key here is that $\arcsin x$ and $-\arccos x$ differ by a constant
$$\arcsin x + C_1 = -\arccos x + \frac{\pi}{2} + C_1 = -\arccos x + C_2$$
What this means is when you take the derivative, the constant goes away, leaving the same function, so the 2 answers you get are equivalent.
