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G(x)=cos(3x) g(x)=G^(')(x) int_(0)^(pi)g(x)dx=

G(x)=cos⁡(3x) G(x)=\cos (3 x) \newlineg(x)=G′(x) g(x)=G^{\prime}(x) \newline∫0πg(x)dx= \int_{0}^{\pi} g(x) d x=

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Full solution

Q. G(x)=cos⁡(3x) G(x)=\cos (3 x) \newlineg(x)=G′(x) g(x)=G^{\prime}(x) \newline∫0πg(x)dx= \int_{0}^{\pi} g(x) d x=
  1. Find g(x)g(x): First, we need to find g(x)g(x) by differentiating G(x)G(x).\newlineG(x)=cos⁡(3x)G(x) = \cos(3x), so g(x)=G′(x)=−3sin⁡(3x)g(x) = G'(x) = -3\sin(3x).
  2. Set up integral: Now, let's set up the integral. ∫0πg(x) dx=∫0π(−3sin⁡(3x)) dx\int_{0}^{\pi}g(x)\,dx = \int_{0}^{\pi}(-3\sin(3x))\,dx.
  3. Integrate −3sin⁡(3x)-3\sin(3x): Integrate −3sin⁡(3x)-3\sin(3x) with respect to xx. The antiderivative of sin⁡(ax)\sin(ax) is −1acos⁡(ax)-\frac{1}{a} \cos(ax), so the antiderivative of −3sin⁡(3x)-3\sin(3x) is −3⋅(−13)⋅cos⁡(3x)=cos⁡(3x)-3 \cdot (-\frac{1}{3}) \cdot \cos(3x) = \cos(3x).
  4. Evaluate integral: Evaluate the integral from 00 to π\pi. \newline∫0π(−3sin⁡(3x)) dx=[cos⁡(3x)]0π=cos⁡(3π)−cos⁡(0).\int_{0}^{\pi}(-3\sin(3x))\,dx = [\cos(3x)]_{0}^{\pi} = \cos(3\pi) - \cos(0).
  5. Calculate values: Calculate the values of cos⁡(3π)\cos(3\pi) and cos⁡(0)\cos(0).cos⁡(3π)=−1\cos(3\pi) = -1 and cos⁡(0)=1\cos(0) = 1.
  6. Substitute values: Substitute these values into the equation. \newlinecos⁡(3π)−cos⁡(0)=−1−1=−2\cos(3\pi) - \cos(0) = -1 - 1 = -2.
  7. Final answer: The final answer is −2-2.

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