Mistake in the computation of $\int_0^1\frac{\mathrm{d}x}{(2-x)\sqrt{1-x}}$ .

Question

The integral:

$$\int_0^1\frac{\mathrm{d}x}{(2-x)\sqrt{1-x}} \color{red}=$$

So I am fond of substitution here:

$$\begin{array}{|ccc|} t = \sqrt{1-x} \\ t^2 = 1 - x \\ 2tdt =-dx \\ -2tdt =dx \\ x = 1 - t^2 \end{array}$$ $$\color{red}= 2\int_0^1 \frac{t\;\mathrm{d}t}{(2-1+t^2)t} = 2\int_0^1\frac{\mathrm{d}t}{1+t^2} = 2\arcsin t \vert_0^1 =2(\arcsin1 - \arcsin0) = \frac{2\pi}{2} = \pi$$

But the final answer is $\frac{\pi}{2}$

What did I miss?

Answer 1

Note that $\int \frac {1}{1+t^2} \ dt=\arctan(t) +c$

Answer 2

$\int\frac{1}{1+t^2}dt=\arctan{t}+C$.

Answer 3

Your mistake was this:

$$\int \frac{1}{1+t^2}dt \neq \arcsin(t) + C$$

But

$$\int \frac{1}{1+t^2}dt = \arctan(t) + C$$