JEE Main 2019MathematicsDefinite IntegrationMediumMCQ

JEE Main 2019 — Definite Integration Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

The value of ∫02πsin⁡2x1+cos⁡3xdx , where [t] denotes the greatest integer function is

Choose an option

Show full solutionCorrect option: C
Correct answer
C-π

Step-by-step explanation

I=∫02πsin2x1+cos3xdx   …1
I=∫02πsin2π-2x1+cos2π-3xdx 

Applying ∫0afx=∫0afa-xdx

I=∫02π-sin2x1+cos3xdx  …2

Adding 1 and 2

2I=∫02πsin⁡2x1+cos⁡3x+-sin⁡2x1+cos⁡3xdx
⇒ 2I=∫02π-1dx   {∵  x+-x=0   :x∈I-1 :x∉I 

⇒ 2I=-x02π

⇒ 2I=-2π

⇒ I= -π

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About this question

This is a previous-year question from JEE Main 2019, 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.