Question

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

Answer 1

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