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प्रश्न
\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
योग
उत्तर
\[\int \left( \frac{\cos^2 x - \sin^2 x}{\sqrt{1 + \cos 4x}} \right)dx\]
`= ∫(\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 ]`
\[ = \frac{1}{\sqrt{2}}\int\left( \frac{\cos 2x}{\cos 2x} \right)dx\]
\[ = \frac{1}{\sqrt{2}}\left[ x \right] + C\]
\[ = \frac{x}{\sqrt{2}} + C\]
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