Inverse trigonometry integration

Question

The magnitude of ans is correct but the sign is negative. Which is incorrect. But the procedure seem to be correct.

Answer 1

The problem lies in your choice of trig identity. Cosine is an even function, so $\cos(\frac{\pi}{2}-x)$=$\cos(x-\frac{\pi}{2})$ The problem with your "cancelation" is that $y=\frac{\pi}{2}-x$ and $y=x-\frac{\pi}{2}$ are not the same graph. On interval $[0,1]$ one is above the axis and one below. That explains your answer off by a negative. So the best thing is to graph the original function first as to determine what trig identity should be used to rewrite the sine.

Answer 2

I think this is because of the property of cos to absorb negative sign of angle.

If you write $\sin x = \cos(\frac π2 - x)$

You got $\frac π2 - \frac12$

Answer 3

The problem is the range of arccosine, which is between $0$ and $\pi$. Therefore, $\cos^{-1}(\cos x)\ne x$ if $x$ is negative.