JEE Main 2022MathematicsDefinite IntegrationEasyMCQ

JEE Main 2022 — Definite Integration Question with Solution

JEE Main 2022 (26 Jul Shift 2)

Question

∫020πsinx+cosx2dx is equal to:

Choose an option

Show full solutionCorrect option: D
Correct answer
D20π+2

Step-by-step explanation

Let I=∫020πsinx+cosx2dx

We know that the period of sinx+cosx is π2

i.e. I=40∫0π2sinx+cosx2dx

I=40∫0π21+sin2xdx

=40x-cos2x20π2=40π2-cosπ2+cos02

I=20π+2

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About this question

This is a previous-year question from JEE Main 2022, covering the Definite 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.