Question

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

Answer 1

\[\int \left( \frac{\cos^2 x - \sin^2 x}{\sqrt{1 + \cos 4x}} \right)dx\]

`= ∫(\text{ cos   ( 2x )} ) / sqrt{2 cos^2 ( 2x ) }  dx          [ ∴ 1 + cos A = 2 cos^2 (A / 2) &    cos^2 A - sin^2 A = cos  2A ]`
\[ = \frac{1}{\sqrt{2}}\int\left( \frac{\cos 2x}{\cos 2x} \right)dx\]
\[ = \frac{1}{\sqrt{2}}\left[ x \right] + C\]
\[ = \frac{x}{\sqrt{2}} + C\]