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Q.

∫cos⁡loge⁡xdx is equal to

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a

12xcos⁡loge⁡x+sin⁡loge⁡x

b

xcos⁡loge⁡x+sin⁡loge⁡x

c

12xcos⁡loge⁡x−sin⁡loge⁡x

d

xcos⁡loge⁡x−sin⁡loge⁡x

answer is A.

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Detailed Solution

I=∫cos⁡loge⁡xdx=∫cos⁡loge⁡x⋅1dx=cos⁡loge⁡xx−∫−sin⁡loge⁡xx⋅xdx=xcos⁡loge⁡x+∫sin⁡loge⁡xdx=xcos⁡loge⁡x+∫sin⁡loge⁡x⋅1dx=xcos⁡loge⁡x+sin⁡loge⁡xx−∫cos⁡loge⁡xxxdx=xcos⁡loge⁡x+xsin⁡loge⁡x−I2I=xcos⁡loge⁡x+sin⁡loge⁡xI=x2cos⁡loge⁡x+sin⁡loge⁡x
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