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Evaluate the integral.

int6x cos(4x)dx
Answer:

Evaluate the integral.\newline6xcos(4x)dx \int 6 x \cos (4 x) d x \newlineAnswer:

Full solution

Q. Evaluate the integral.\newline6xcos(4x)dx \int 6 x \cos (4 x) d x \newlineAnswer:
  1. Identify integral: Identify the integral to be solved.\newlineWe need to evaluate the integral of the function 6xcos(4x)6x \cos(4x) with respect to xx, which is written as 6xcos(4x)dx\int 6x \cos(4x) \, dx.
  2. Use integration by parts: Use integration by parts.\newlineIntegration by parts is given by the formula udv=uvvdu\int u \, dv = uv - \int v \, du, where uu and dvdv are parts of the integrand. We choose uu and dvdv as follows:\newlineLet u=6xu = 6x, which implies du=6dxdu = 6 \, dx.\newlineLet dv=cos(4x)dxdv = \cos(4x) \, dx, which implies v=14sin(4x)v = \frac{1}{4}\sin(4x) after integrating with respect to xx.
  3. Apply integration by parts: Apply the integration by parts formula.\newlineUsing the chosen uu and dvdv, we have:\newline6xcos(4x)dx=uvvdu\int 6x \cos(4x) \, dx = uv - \int v \, du\newline=(6x)(14)sin(4x)(14)sin(4x)(6dx)= (6x)(\frac{1}{4})\sin(4x) - \int(\frac{1}{4})\sin(4x)(6 \, dx)\newline=(32)xsin(4x)(32)sin(4x)dx= (\frac{3}{2})x \sin(4x) - (\frac{3}{2})\int\sin(4x) \, dx
  4. Integrate remaining integral: Integrate the remaining integral.\newlineNow we need to integrate sin(4x)dx\int \sin(4x) \, dx. The antiderivative of sin(4x)\sin(4x) with respect to xx is 14cos(4x)-\frac{1}{4}\cos(4x). So we have:\newline\frac{3}{2}\int \sin(4x) \, dx = \frac{3}{2}\left(-\frac{1}{4}\cos(4x)\right) + C\(\newline= -\frac{3}{8}\cos(4x) + C\), where CC is the constant of integration.
  5. Combine final answer: Combine the results to get the final answer.\newlinePutting it all together, we have:\newline6xcos(4x)dx=(32)xsin(4x)(38)cos(4x)+C\int 6x \cos(4x) \, dx = \left(\frac{3}{2}\right)x \sin(4x) - \left(\frac{3}{8}\right)\cos(4x) + C