Find the following antiderivative: $\int{\frac{1}{\sqrt{(2-x)(x-1)}}}dx$
My attempt:
$$\int{\frac{1}{\sqrt{(2-x)(x-1)}}}dx$$
$$=\int{\frac{1}{\sqrt{2x-2-x^2+x}}}dx$$
$$=\int{\frac{1}{\sqrt{-2+3x-x^2}}}dx$$
$$=\int{\frac{1}{\sqrt{-(2-3x+x^2)}}}dx$$
$$=\int{\frac{1}{\sqrt{-(x^2-3x+2)}}}dx$$
$$=\int{\frac{1}{\sqrt{-(x^2-3x+{(\frac{3}{2})^2)}-2+(\frac{3}{2})^2}}}dx$$
$$=\int{\frac{1}{\sqrt{0.5^2-(x-\frac{3}{2})^2}}}dx$$
$$[\text{Let $x-\frac{3}{2}=u$}\\ \therefore du=dx]$$
$$=\int{\frac{1}{\sqrt{0.5^2-u^2}}}dx$$
$$[\text{Formula:}\int{\frac{1}{\sqrt{a^2-x^2}}}dx=\arcsin\left(\frac{x}{a}\right)+C, |x|<a]$$
$$=\arcsin\left(\frac{u}{0.5}\right)+C$$
$$=\arcsin\left(\frac{x-\frac{3}{2}}{0.5}\right)+C$$
$$=\arcsin(2x-3)+C$$
My work is correct. I know this because integral-calculator agrees with me (after going to the link, click on "Go!". For steps, click on "Show steps").
However, my book did this math in a different way:
My book's attempt:
$$\int{\frac{1}{\sqrt{(2-x)(x-1)}}}dx$$
$$[\text{Let}\ x-1=u^2,dx=2udu,x=u^2+1,2-x=1-u^2]$$
$$=\int{\frac{1}{\sqrt{(1-u^2)(u^2)}}}2udu$$
$$=\int{\frac{1}{\sqrt{(1-u^2)}u}}2udu$$
$$=2\int{\frac{1}{\sqrt{(1-u^2)}}}du$$
$$[\text{Formula:}\int{\frac{1}{\sqrt{a^2-x^2}}}dx=\arcsin\left(\frac{x}{a}\right)+C, |x|<a]$$
$$=2\arcsin(u)+C$$
$$=2\arcsin(x-1)+C$$
Comments:
This is the graph. From the graph, it doesn't look like the book's answer differs only by a constant from my answer. So, is it wrong? If it is wrong, what mistakes did it make?
Best Answer
The very last equality looks dodgy to me, $u\ne x-1$.