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Question: Answered & Verified by Expert
The integral I=1xsecx-lnxsinxdx simplifies to (where, c is the constant of integration)
MathematicsIndefinite IntegrationJEE Main
Options:
  • A lnxsinx+c
  • B lnxcosx+c
  • C xlnsinx+c
  • D xlncosx+c
Solution:
2750 Upvotes Verified Answer
The correct answer is: lnxcosx+c
Given I=cosx1x+-sinxlnxdx
Since,fg'+f'gdx
=fg+c
I=cosxlnx+c

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