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The value of secxdxsin(2x+θ)+sinθ is 

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a
(tan⁡x+tan⁡θ)sec⁡θ+C
b
2(tan⁡x+tan⁡θ)sec⁡θ+C
c
2(sin⁡x+tan⁡θ)sec⁡θ+C
d
none of these

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detailed solution

Correct option is B

The given integrals can be written as ∫sec⁡xdx2sin⁡(x+θ)cos⁡x=12∫sec3/2⁡xdxsin⁡xcos⁡θ+sin⁡θcos⁡x=12∫sec2⁡xdxtan⁡xcos⁡θ+sin⁡θ=12∫dttcos⁡θ+sin⁡θ=22tcos⁡θ+sin⁡θ+C=2(tan⁡xsec⁡θ+tan⁡θsec⁡θ)+C


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