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∫ Sin − 1 ( 2 Tan X 1 + Tan 2 X ) D X - Mathematics

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प्रश्न

\[\int \sin^{- 1} \left( \frac{2 \tan x}{1 + \tan^2 x} \right) dx\]
बेरीज
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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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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 38 | पृष्ठ १५

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