Question:
The value of \(\displaystyle\int_0^{2\pi} \left[\sin 2x\,(1 + \cos 3x)\right] dx\), where \([t]\) denotes the greatest integer function is :
The value of \(\displaystyle\int_0^{2\pi} \left[\sin 2x\,(1 + \cos 3x)\right] dx\), where \([t]\) denotes the greatest integer function is :
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Let \(f(x)=\sin2x(1+\cos3x)\). Then \(f(2\pi-x)=-f(x)\), and \([t]+[-t]=-1\) for non-integer \(t\); use \(\int_0^{2a}h(x)dx=\int_0^{2a}h(2a-x)dx\).
Updated On: Nov 5, 2026
- \(2\pi\)
- \(\pi\)
- \(-2\pi\)
- \(-\pi\)
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The Correct Option is D
Solution and Explanation
Let \(f(x) = \sin2x(1+\cos3x)\) and \(I = \int_0^{2\pi}[f(x)]\,dx\).
Replace \(x\to 2\pi-x\): \(\sin 2(2\pi-x) = -\sin 2x\) and \(\cos3(2\pi-x)=\cos3x\), so \(f(2\pi-x) = -f(x)\).
Hence \(I = \int_0^{2\pi}[f(2\pi-x)]\,dx = \int_0^{2\pi}[-f(x)]\,dx\).
Adding: \(2I = \int_0^{2\pi}\big([f(x)]+[-f(x)]\big)dx\).
For non-integer \(t\), \([t]+[-t]=-1\); \(f(x)\) is an integer only at finitely many points, which do not affect the integral.
So \(2I = \int_0^{2\pi}(-1)\,dx = -2\pi\) and \(I = -\pi\).
Answer: (4) \(-\pi\)
Replace \(x\to 2\pi-x\): \(\sin 2(2\pi-x) = -\sin 2x\) and \(\cos3(2\pi-x)=\cos3x\), so \(f(2\pi-x) = -f(x)\).
Hence \(I = \int_0^{2\pi}[f(2\pi-x)]\,dx = \int_0^{2\pi}[-f(x)]\,dx\).
Adding: \(2I = \int_0^{2\pi}\big([f(x)]+[-f(x)]\big)dx\).
For non-integer \(t\), \([t]+[-t]=-1\); \(f(x)\) is an integer only at finitely many points, which do not affect the integral.
So \(2I = \int_0^{2\pi}(-1)\,dx = -2\pi\) and \(I = -\pi\).
Answer: (4) \(-\pi\)
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