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$I = \int \cos \{2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}\} d x$ is equal to

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Correct option is A

Put x=cos⁡θ⇒dx=−sin⁡θdθI=∫cos⁡2tan−1⁡tan⁡θ/2(−sin⁡θ)dθ=−∫cos⁡θsin⁡θdθ=cos2⁡θ2+C=x22+C

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I=∫cos⁡2tan−1⁡1−x1+xdx is equal to

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$I = \int \cos \{2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}\} d x$ is equal to