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Evaluate the integral. int-2x^(3)cos(3x)dx Answer:

Evaluate the integral.\newline∫−2x3cos⁡(3x)dx \int-2 x^{3} \cos (3 x) d x \newlineAnswer:

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Q. Evaluate the integral.\newline∫−2x3cos⁡(3x)dx \int-2 x^{3} \cos (3 x) d x \newlineAnswer:
  1. Identify Integral: Identify the integral that needs to be evaluated.\newlineWe need to evaluate the integral of the function −2x3cos⁡(3x)-2x^3 \cos(3x) with respect to xx. This is an integration problem that requires the use of integration by parts, which is based on the formula ∫u dv=uv−∫v du\int u \, dv = uv - \int v \, du.
  2. Choose uu and dvdv: Choose uu and dvdv for integration by parts.\newlineLet u=−2x3u = -2x^3, which means du=−6x2dxdu = -6x^2 dx after differentiation.\newlineLet dv=cos⁡(3x)dxdv = \cos(3x) dx, which means v=(1/3)sin⁡(3x)v = (1/3)\sin(3x) after integration.
  3. Apply Integration by Parts: Apply the integration by parts formula.\newlineUsing the integration by parts formula, we have:\newline∫−2x3cos⁡(3x) dx=uv−∫v du\int -2x^3 \cos(3x) \, dx = uv - \int v \, du\newline= (−2x3)(13)sin⁡(3x)−∫(13)sin⁡(3x)(−6x2 dx)(-2x^3)(\frac{1}{3})\sin(3x) - \int (\frac{1}{3})\sin(3x)(-6x^2 \, dx)\newline= (-\frac{\(2\)}{\(3\)})x^\(3 \sin(33x) + 22 \int x^22 \sin(33x) \, dx)
  4. Apply Integration by Parts Again: Apply integration by parts again to the remaining integral.\newlineWe need to apply integration by parts again to the integral ∫x2sin⁡(3x) dx\int x^2 \sin(3x) \, dx.\newlineLet u=x2u = x^2, which means du=2x dxdu = 2x \, dx after differentiation.\newlineLet dv=sin⁡(3x) dxdv = \sin(3x) \, dx, which means v=−(1/3)cos⁡(3x)v = -(1/3)\cos(3x) after integration.
  5. Apply Integration by Parts Formula: Apply the integration by parts formula to the new integral.\newlineUsing the integration by parts formula again, we have:\newline∫x2sin⁡(3x) dx=uv−∫v du\int x^2 \sin(3x) \, dx = uv - \int v \, du\newline=x2(−13)cos⁡(3x)−∫(−13)cos⁡(3x)(2x dx)= x^2(-\frac{1}{3})\cos(3x) - \int(-\frac{1}{3})\cos(3x)(2x \, dx)\newline=−(13)x2cos⁡(3x)+(23)∫xcos⁡(3x) dx= -(\frac{1}{3})x^2 \cos(3x) + (\frac{2}{3}) \int x \cos(3x) \, dx
  6. Apply Integration by Parts Again: Apply integration by parts one more time to the remaining integral.\newlineWe need to apply integration by parts one more time to the integral ∫xcos⁡(3x) dx\int x \cos(3x) \, dx.\newlineLet u=xu = x, which means du=dxdu = dx after differentiation.\newlineLet dv=cos⁡(3x) dxdv = \cos(3x) \, dx, which means v=(13)sin⁡(3x)v = (\frac{1}{3})\sin(3x) after integration.
  7. Apply Integration by Parts Formula: Apply the integration by parts formula to the final integral.\newlineUsing the integration by parts formula one more time, we have:\newline∫xcos⁡(3x) dx=uv−∫v du\int x \cos(3x) \, dx = uv - \int v \, du\newline= x(13)sin⁡(3x)−∫(13)sin⁡(3x) dxx(\frac{1}{3})\sin(3x) - \int(\frac{1}{3})\sin(3x) \, dx\newline= (13)xsin⁡(3x)−(19)cos⁡(3x)+C(\frac{1}{3})x \sin(3x) - (\frac{1}{9})\cos(3x) + C
  8. Combine for Final Answer: Combine all parts to get the final answer.\newlineNow we combine all parts to get the final answer for the original integral:\newline∫−2x3cos⁡(3x) dx=(−23)x3sin⁡(3x)+2[−(13)x2cos⁡(3x)+(23)((13)xsin⁡(3x)−(19)cos⁡(3x))]+C\int -2x^3 \cos(3x) \, dx = \left(-\frac{2}{3}\right)x^3 \sin(3x) + 2\left[-\left(\frac{1}{3}\right)x^2 \cos(3x) + \left(\frac{2}{3}\right)\left(\left(\frac{1}{3}\right)x \sin(3x) - \left(\frac{1}{9}\right)\cos(3x)\right)\right] + C\newlineSimplify the expression:\newline=(−23)x3sin⁡(3x)−(23)x2cos⁡(3x)+(49)xsin⁡(3x)−(427)cos⁡(3x)+C= \left(-\frac{2}{3}\right)x^3 \sin(3x) - \left(\frac{2}{3}\right)x^2 \cos(3x) + \left(\frac{4}{9}\right)x \sin(3x) - \left(\frac{4}{27}\right)\cos(3x) + C

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