∫sin2⁡x⋅sec2⁡x+2tan⁡x⋅sin−1⁡x⋅1−x21−x21+tan2⁡xdx=

∫sin2⁡x⋅sec2⁡x+2tan⁡x⋅sin−1⁡x⋅1−x21−x21+tan2⁡xdx=∫sin2⁡x1−x2+2tan⁡x⋅sin−1⁡xsec2⁡xdx=∫sin2⁡x1−x2dx+∫2sin⁡xcos⁡xsin−1⁡xdx =sin2⁡xsin−1⁡x−∫2sin⁡xcos⁡xsin−1⁡xdx+∫2sin⁡xcos⁡xsin−1⁡xdx+C=sin2⁡xsin−1⁡x+C

∫sin2⁡x⋅sec2⁡x+2tan⁡x⋅sin−1⁡x⋅1−x21−x21+tan2⁡xdx=

∫sin2⁡x⋅sec2⁡x+2tan⁡x⋅sin−1⁡x⋅1−x21−x21+tan2⁡xdx=∫sin2⁡x1−x2+2tan⁡x⋅sin−1⁡xsec2⁡xdx=∫sin2⁡x1−x2dx+∫2sin⁡xcos⁡xsin−1⁡xdx =sin2⁡xsin−1⁡x−∫2sin⁡xcos⁡xsin−1⁡xdx+∫2sin⁡xcos⁡xsin−1⁡xdx+C=sin2⁡xsin−1⁡x+C

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$\int \frac{\sin^{2} x \cdot \sec^{2} x + 2 \tan x \cdot \sin^{- 1} x \cdot \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}} (1 + \tan^{2} x)} d x =$

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$(\sec^{2} x) \cdot (\cos^{- 1} x) + c$

$(\sin^{2} x) \cdot (\sin^{- 1} x) + c$

$(\sec^{2} x) \cdot (\tan^{- 1} x) + c$

$(\cos^{2} x) \cdot (\sin^{- 1} x) + c$

answer is B.

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Detailed Solution

$\begin{array}{l} \int \frac{\sin^{2} x \cdot \sec^{2} x + 2 \tan x \cdot \sin^{- 1} x \cdot \sqrt{1 - x^{2}}}{\sqrt{1 - x^{2}} (1 + \tan^{2} x)} d x \\ = \int (\frac{\sin^{2} x}{\sqrt{1 - x^{2}}} + \frac{2 \tan x \cdot \sin^{- 1} x}{\sec^{2} x}) d x \\ = \int \frac{\sin^{2} x}{\sqrt{1 - x^{2}}} d x + \int 2 \sin x \cos x \sin^{- 1} x d x \begin{array}{l} = \sin^{2} x \sin^{- 1} x - \int 2 \sin x \cos {xsin}^{- 1} x d x + \int 2 \sin x \cos x \sin^{- 1} x d x + C \\ = \sin^{2} x \sin^{- 1} x + C \end{array} \end{array}$