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