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प्रश्न
\[\int\frac{1 - \cos 2x}{1 + \cos 2x} dx\]
बेरीज
उत्तर
\[\int\left( \frac{1 - \cos 2x}{1 + \cos 2x} \right)dx\]
`=∫ (2 sin^2 x) / ( 2 cos^2 x ) dx [∵ 1 - cos 2 θ = 2 sin^2 θ & 1 + cos 2 θ= 2 cos^2 θ]`
\[ = \int \tan^2 \text{x dx} \]
\[ = \int\left( \sec^2 x - 1 \right) dx\]
\[ = \int \sec^2\text{ x dx} - ∫ dx\]
\[ = \tan x - x + C\]
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