JEE Main 2020MathematicsIndefinite IntegrationHardMCQ

JEE Main 2020 — Indefinite Integration Question with Solution

JEE Main 2020 (04 Sep Shift 1)

Question

The integral ∫xxsinx+cosx2dx is equal to, (where C is a constant of integration):

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Show full solutionCorrect option: A
Correct answer
Atanx-xsecxxsinx+cosx+C

Step-by-step explanation

∫x2    dxxsinx+cosx2=∫xcosx⋅xcosxxsinx+cosx2 dx

=xcosx∫xcosxxsinx+cosx2dx−∫ddxxsecx∫xcosxxsinx+cosx2 dx dx

=xcosx−1xsinx+cosx+∫sec2x  dx, ∵    ∫xcosxxsinx+cosx2dx=−1xsinx+cosx

 & ddxxsecx=secx+xsecxtanx=secx1+xsinxcosx=sec2xxsinx+cosx

=−xcosxxsinx+cosx+sinxcosx+C

=tanx−xsecxxsinx+cosx+C

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About this question

This is a previous-year question from JEE Main 2020, covering the Indefinite Integration chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.