Question

$\int \frac{{\cos }^{2}x-{\sin }^{2}x}{\sqrt{1}+\cos 4x}dx$

Answer 1

$$\begin{gathered} \int \left(\frac{{\cos }^{2}x-{\sin }^{2}x}{\sqrt{1+\cos 4x}}\right)dx \\ =\int \frac{\cos \left(2x\right)}{\sqrt{2{\cos }^2\left(2x\right)}}dx\left[\therefore 1+\cos \mathrm{A}=2{\cos }^2\left(\frac{\mathrm{A}}{2}\right)\&{\cos }^2\mathrm{A}-{\sin }^2\mathrm{A}=\cos 2\mathrm{A}\right] \\ =\frac{1}{\sqrt{2}}\int \left(\frac{\cos 2x}{\cos 2x}\right)dx \\ =\frac{1}{\sqrt{2}}\left[x\right]+\mathrm{C} \\ =\frac{x}{\sqrt{2}}+\mathrm{C} \end{gathered}$$