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int_(2pi)^(4pi)sin^(2)xdx

∫2π4πsin⁡2xdx \int_{2 \pi}^{4 \pi} \sin ^{2} x d x

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Full solution

Q. ∫2π4πsin⁡2xdx \int_{2 \pi}^{4 \pi} \sin ^{2} x d x
  1. Split Integral: Now, let's split the integral into two parts. ∫2π4π1−cos⁡(2x)2dx=12∫2π4πdx−12∫2π4πcos⁡(2x)dx\int_{2\pi}^{4\pi}\frac{1 - \cos(2x)}{2} dx = \frac{1}{2} \int_{2\pi}^{4\pi}dx - \frac{1}{2} \int_{2\pi}^{4\pi}\cos(2x)dx
  2. Calculate First Integral: Calculate the first integral, which is a constant.\newline12∫2π4πdx=12[x]2π4π=12[4π−2π]=12[2π]=π\frac{1}{2} \int_{2\pi}^{4\pi}dx = \frac{1}{2} [x]_{2\pi}^{4\pi} = \frac{1}{2} [4\pi - 2\pi] = \frac{1}{2} [2\pi] = \pi
  3. Calculate Second Integral: Now, calculate the second integral, which involves the cosine function.\newline12∫2π4πcos⁡(2x) dx=12[sin⁡(2x)2]2π4π\frac{1}{2} \int_{2\pi}^{4\pi}\cos(2x)\,dx = \frac{1}{2} \left[\frac{\sin(2x)}{2}\right]_{2\pi}^{4\pi}
  4. Plug in Limits: Plug in the limits for the second integral.\newline12[sin⁡(2x)2]2π4π=12[sin⁡(8π)2−sin⁡(4π)2]\frac{1}{2} \left[\frac{\sin(2x)}{2}\right]_{2\pi}^{4\pi} = \frac{1}{2} \left[\frac{\sin(8\pi)}{2} - \frac{\sin(4\pi)}{2}\right]\newlineSince sin⁡(8π)=sin⁡(4π)=0\sin(8\pi) = \sin(4\pi) = 0, this part of the integral is zero.
  5. Add Results: Add the results of the two integrals. π+0=π\pi + 0 = \pi So, the final answer is π\pi.

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