Question

$\int \sqrt{2x-x^2}dx$

Answer 1

$$\begin{gathered} I=\int \sqrt{2x-x^2}dx \\ =\int \sqrt{x\left(2-x\right)}dx \end{gathered}$$

Let $x=1+\sin u$

$$\begin{gathered} \mathrm{or},dx=\cos udu \\ \Rightarrow I=\int \sqrt{\left(1+\sin u\right)\left(1-\sin u\right)}\cos udu \\ \Rightarrow I=\int {\cos }^{2}udu \\ \Rightarrow I=\frac{1}{2}\int \left(\cos 2u+1\right)du \end{gathered}$$

$$\begin{gathered} \Rightarrow I=\frac{1}{2}\left(\frac{1}{2}\sin 2u+u\right)+c \\ \Rightarrow I=\frac{1}{2}\left(\sin u\cos u+u\right)+c \\ \Rightarrow I=\frac{1}{2}\left(\sin u\sqrt{1-{\sin }^{2}u}+u\right)+c \\ \therefore I=\frac{1}{2}\left(x-1\right)\sqrt{2x-x^2}+\frac{1}{2}{\sin }^{-1}\left(x-1\right)+c\left[\because u={\sin }^{-1}\left(x-1\right)\right] \end{gathered}$$