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प्रश्न
\[\int \tan^{- 1} \left( \frac{\sin 2x}{1 + \cos 2x} \right) dx\]
योग
उत्तर
\[\int \tan^{- 1} \left[ \frac{\sin \left( 2x \right)}{1 + \cos2x} \right]dx\]
`= ∫ tan ^-1 [ (2 sin x cos x) / ( 2 cos^2 x)] `dx ` [∴ sin 2x = 2 sin x cos x & 1 + cos 2x = 2 cos ^2 x ]`
\[ = \int \tan^{- 1} \left[ \tan x \right]\]
\[ = \int \tan^{- 1} \left[ \tan x \right]\]
` = ∫ x dx `
\[ = \frac{x^2}{2} + C\]
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