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coslogexdx 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

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detailed solution

Correct option is A

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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