How to calculate the integral $\int_{0.5}^{1} \frac{1}{\sqrt{2x-x^2}}$?

Question

I have the following integration problem:

$$ \int_{0.5}^{1} \frac{1}{\sqrt{2x-x^2}} $$

And I can see I should probably be completing the square here. I may be missing something extremely obvious, but wouldn't this mean I have to take the negative outside the expression to do this? I need the $x^2$ term by itself, don't I?

Any help would be greatly appreciated!

Answer 1

Given $\int_{0.5}^{1}\dfrac{1}{\sqrt{2x-x^2}}dx$

$$=\int\frac{1}{\sqrt{1-(x-1)^2}}dx$$ Apply $u$ substitution $u=x-1$ $$\int\frac{1}{\sqrt{1-u^2}}du=\arcsin(u)=\arcsin(x-1)$$ Now apply the boundaries $$=\Bigl[\arcsin (x-1)\Bigr]_{0.5}^1=\frac{\pi}{6}$$

Answer 2

Note that $$2x - x^2 = 1 - (1 - 2x + x^2) = 1 - (1-x)^2 \, .$$ Now you can use a trig substitution.

Answer 3

There's no hyperbolic function here.

Set $$ t=\sqrt{\frac{x}{2-x}} $$ so $(2-x)t^2=x$ and $$ x=\frac{2t^2}{1+t^2}=\frac{2t^2+2-2}{1+t^2}=2-\frac{2}{1+t^2} $$ whence $$ dx=\frac{4t}{(1+t^2)^2} $$ and the integral becomes \begin{align} \int_{1/2}^1\frac{1}{2x-x^2}\,dx &=\int_{1/2}^1\frac{1}{x}\sqrt{\frac{x}{2-x}}\,dx \\[4px] &=\int_{1/\sqrt{3}}^{1}\frac{1+t^2}{2t^2}t\frac{4t}{(1+t^2)^2}\,dt \\[4px] &=2\int_{1/\sqrt{3}}^1\frac{1}{1+t^2}\,dt \\[4px] &=2\Bigl[\arctan t\Bigr]_{1/\sqrt{3}}^1 \\[4px] &=2\frac{\pi}{4}-2\frac{\pi}{6} \\[4px] &=\frac{\pi}{6} \end{align}