How to arrive at $\sin\theta + c$ from $\int {\mathrm{dx}\over x\sqrt{x^2-2x}}$?

Question

$$\int {\mathrm{dx}\over x\sqrt{x^2-2x}}$$ transform $x^2-2x$:$$\int {\mathrm{dx}\over x\sqrt{(x-1)^2-1}}$$ substituting x with u:$$\int {\mathrm{du}\over (u+1)\sqrt{u^2-1}}$$ substituting u with $\sec\theta$ $$\int {\sec\theta\tan\theta\mathrm{d\theta}\over (\sec\theta+1)\sqrt{\sec^2\theta-1}}$$ simplifying:$$\int {\sec\theta\tan\theta\mathrm{d\theta}\over (\sec\theta+1)\sqrt{\sec^2\theta-1}}$$ $$\int {\sec\theta\tan\theta\mathrm{d\theta}\over (\sec\theta+1)\sqrt{\tan^2\theta}}$$ $$\int {\sec\theta\tan\theta\mathrm{d\theta}\over (\sec\theta+1)\tan\theta}$$ $$\int{\sec\theta\over\sec\theta+1}\;\mathrm{d\theta}$$

I don't know how to manipulate the function to be: $$\int{\cos\theta}\mathrm{d\theta}$$ or maybe I made a mistake somewhere?

Answer 1

Why not just integrate directly

$$\int \frac{dx}{x\sqrt{x^2-2x}}=-\frac12 \int \frac{d(\frac2x)}{\sqrt{1-\frac2x}}=\sqrt{1-\frac2x}+C $$ or, force the substitution $\cos^2\theta = \frac2x$.

Answer 2

Note that$$\int\frac{\sec\theta\:d\theta}{\sec\theta+1}=\int\frac{d\theta}{1+\cos\theta}=\int\frac12\sec^2\frac{\theta}{2}\:d\theta=\tan\frac{\theta}{2}+C.$$But$$t:=\tan\frac{\theta}{2}\implies u=\frac{1+t^2}{1-t^2}\implies t=\pm\sqrt{\frac{1-u}{1+u}}.$$If you want a definition of $\theta$ that gets the antiderivatives as $\sin\theta+C$, @Quanto has noted $x=2\sec^2\theta$ suffices.