Correct option is 3Tanx−Secx+C

Questions

$\int \frac{1}{1 + \sin x} d x .$

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Secx+Tanx+C

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Tanx−Secx+C

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detailed solution

Correct option is C

∫11+sin⁡xdx=∫1(1+sin⁡x)⋅(1−sin⁡x)(1−sin⁡x)dx=∫1−sin⁡x1−sin2⁡xdx==∫1−sin⁡xcos2⁡xdx=∫1cos2⁡xdx−∫sin⁡xcos2⁡xdx=∫sec2⁡xdx−∫tan⁡xsec⁡xdx=tan⁡x−sec⁡x+C

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∫11+sinxdx.

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$\int \frac{1}{1 + \sin x} d x .$