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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?

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    $\begingroup$ $\int \frac{1}{1+x^2} \ dx = \tan^{-1}(x) +C.$ You incorrectly said the antiderivative was $\sin^{-1}(x)$ $\endgroup$ – Vivek Kaushik Jun 1 '17 at 15:33
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    $\begingroup$ $\int \frac{dt}{1+t^2}$ is $\arctan$, no? $\endgroup$ – sharding4 Jun 1 '17 at 15:33
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    $\begingroup$ Indeed, overworked, it is a pity I had to write so much latex code for table integral. Maybe the integral itself will be useful for others, wil keep it $\endgroup$ – M.Mass Jun 1 '17 at 15:41
  • $\begingroup$ [+1] All efforts, sooner or later, are rewarded... Taking time to have a well written (latex and english) question is appreciated by everybody! $\endgroup$ – Jean Marie Jun 1 '17 at 15:50
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Note that $\int \frac {1}{1+t^2} \ dt=\arctan(t) +c$

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

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Your mistake was this:

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

But

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

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