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Q.

∫cos⁡2x−cos⁡2θcos⁡x−cos⁡θdx is equal to

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a

2(sin⁡x+xcos⁡θ)+C

b

2(sin⁡x-xcos⁡θ)+C

c

2(sin⁡x+2xcos⁡θ)+C

d

2(sin⁡x−2xcos⁡θ)+C

answer is A.

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Detailed Solution

∫cos⁡2x−cos⁡2θcos⁡x−cos⁡θdθ=∫2cos2⁡x−1−2cos2⁡θ−1cos⁡x−cos⁡θdθ                                   =2∫cos2⁡x−cos2⁡θcos⁡x−cos⁡θdθ=2∫(cos⁡x+cos⁡θ)dθ=2sin⁡x+2xcos⁡θ+C
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∫cos⁡2x−cos⁡2θcos⁡x−cos⁡θdx is equal to