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Q.

∫11+sinxdx.

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a

Secx+Tanx+C

b

Secx−Tanx+C

c

Tanx−Secx+C

d

Tanx-Secx+C

answer is C.

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Detailed Solution

∫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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