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Question:

The value of (\int_0^{2\pi} \sin(2x)(1+\cos(3x))dx) is:

-2;π

π

Solution:

I=\int_0^{2\pi} \sin(2x)(1+\cos(3x))dx
=\int_0^{2\pi} (\sin(2x) + \sin(2x)\cos(3x))dx
=\int_0^{2\pi} \sin(2x)dx + \int_0^{2\pi} \sin(2x)\cos(3x)dx
=\left[ -\frac{\cos(2x)}{2} \right]_0^{2\pi} + \frac{1}{2}\int_0^{2\pi} (\sin(5x) - \sin(x))dx
=\left( -\frac{1}{2} + \frac{1}{2} \right) + \frac{1}{2} \left[ -\frac{\cos(5x)}{5} + \cos(x) \right]_0^{2\pi}
=0 + \frac{1}{2} \left( -\frac{1}{5} + 1 + \frac{1}{5} - 1 \right)
=0