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Can someone explain - Integral Calculus - JEE Main-17

The value of \int_{0}^{2\pi}[sin2x(1+cos3x)]dx, where [t] denotes

greatest integer function, is :

  • Option 1)

    \pi

  • Option 2)

    -\pi

  • Option 3)

    -2\pi

  • Option 4)

    2\pi

 
Answers (1)
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V Vakul

I= \int_{0}^{2\pi}[sin2x(1+cos3x)]dx.....................(1)

I= \int_{0}^{2\pi}[sin2(2\pi-x)(1+cos3(2\pi-x))]dx

I= \int_{0}^{2\pi}[-sin2x(1+cos3x)]dx..............................(2)

Add (1) and (2)

2I= \int_{0}^{2\pi}([sin2x(1+cos3x)]+[-sin2x(cos 3x+1)])dx

2I= \int_{0}^{2\pi}-1dx

2I= [-x]_{0}^{2\pi}

2I= -2\pi

I= -\pi

So, option (2) is correct.


Option 1)

\pi

Option 2)

-\pi

Option 3)

-2\pi

Option 4)

2\pi

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