JEE Main 2018MathematicsIndefinite IntegrationMediumMCQ

JEE Main 2018 — Indefinite Integration Question with Solution

JEE Main 2018 (08 Apr)

Question

The integral ∫sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2dx, is equal to

(where C is the constant of integration).

Choose an option

Show full solutionCorrect option: C
Correct answer
C-131+tan3x+C

Step-by-step explanation

∫sin2xcos2xsin5x+cos3xsin2x+sin3xcos2x+cos5x2dx

=∫sin2xcos2xsin3xsin2x+cos2x+cos3x cos2x+sin2x2dx 

=∫sin2xcos2xsin3x+cos3x2dx

=∫sin2xcos2xcos3x2tan3x+12dx

=∫tan2xsec2xtan3x+12dx
Put, tan3x+1=t
⇒dt=3tan2xsec2xdx
=∫dt3t2=-13t+C, where C is the constant of integration.
=  -13tan3x+1+C

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About this question

This is a previous-year question from JEE Main 2018, 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.