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∫ Cos 2 X − Sin 2 X √ 1 + Cos 4 X D X - Mathematics

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प्रश्न

\[\int\frac{\cos^2 x - \sin^2 x}{\sqrt{1} + \cos 4x} dx\]
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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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पाठ 19: Indefinite Integrals - Exercise 19.02 [पृष्ठ १५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.02 | Q 28 | पृष्ठ १५

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