Question

$\int \sin x\sqrt{1-\cos 2x}dx$

Answer 1

$$\begin{gathered} \int \sin x·\sqrt{1-\cos 2x}dx \\ =\int \sin x.\sqrt{2{\sin }^{2}x}dx\left[\therefore 1-\cos 2\mathrm{A}=2{\sin }^2\mathrm{A}\right] \\ =\sqrt{2}\int {\sin }^{2}xdx \\ =\sqrt{2}\int \left(\frac{1-\cos 2x}{2}\right)dx \\ =\frac{1}{\sqrt{2}}\int \left(1-\cos 2x\right)dx \\ =\frac{1}{\sqrt{2}}\left[x-\frac{\sin 2x}{2}\right]+\mathrm{C} \end{gathered}$$