Use the methods introduced evaluate the following integrals. ∫x2cos⁡xdx

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Use the methods introduced evaluate the following integrals.  ∫x2cos⁡xdx

MoxboasteBots5h · Accepted Answer

Step 1Given the integral ∫x2cos⁡xdx.We need to evaluate the above integral.Step 2Let I=∫(x2cos⁡x)dxWe need to integrate the above integral using integrating by parts.u=x2 and v=cos⁡xSo ∫(uv)dx=u∫vdx−∫[u′∫vdx]I=x2sin⁡x−∫2xsin⁡xdxI=x2sin⁡x−2∫(xsin⁡x)dxwhere∫(x sin⁡x)dx=x(−cos⁡x)−∫1⋅(cos⁡x)dx=−xcos⁡x+∫cos⁡xdx=−xcos⁡x+sin⁡xI=x2sin⁡x−2[−xcos⁡x+sin⁡x]+CI=x2sin⁡x+2xcos⁡x−2sin⁡x+C

Mary Herrera · Accepted Answer

∫x2⋅cos⁡(x)dx  Prepare for integration by parts  u=x2  dv=cos⁡(x)dx  Calculate the differential  du=2xdx  v=sin⁡(x)  Substitute the values into the formula  x2⋅sin⁡(x)−∫sin⁡(x)⋅2xdx  x2⋅sin⁡(x)−2⋅∫sin⁡(x)⋅xdx  x2⋅sin⁡(x)−2⋅∫x⋅sin⁡(x)dx  Use the partial integration  x2⋅sin⁡(x)−2(x⋅(−cos⁡(x))−∫−cos⁡(x)dx)  x2⋅sin⁡(x)−2(x⋅(−cos⁡(x))+∫cos⁡(x)dx)  x2⋅sin⁡(x)−2(x⋅(−cos⁡(x))+sin⁡(x))  Simplify  x2⋅sin⁡(x)+2x⋅cos⁡(x)−2sin⁡(x)  Answer:  x2⋅sin⁡(x)+2x⋅cos⁡(x)−2sin⁡(x)+C

Vasquez · Accepted Answer

∫x2cos⁡(x)dxIntegration formula by parts:∫U(x)∗dV(x)=U(x)∗V(x)−∫V(x)∗dU(x)PutU=x2dV=cos⁡(x)dxThen:dU=2xdxV=sin⁡(x)Therefore:∫x2∗cos⁡(x)dx=x2∗sin⁡(x)−∫2x∗sin⁡(x)dxFind the integral∫2x∗sin⁡(x)dxU=xdU=dxdV=2sin⁡(x)dxV=−2cos⁡(x)∫2x∗sin⁡(x)dx=−2x∗cos⁡(x)dx=−2x∗cos⁡(x)+∫2∗cos⁡(x)dxFind the integral∫2∗cos⁡(x)dx=2∗sin⁡(x)Answer:∫x2∗cos⁡(x)=x2∗sin⁡(x)+2x∗cos⁡(x)−2∗sin⁡(x)+C

Use the methods introduced evaluate the following integrals. \int x^{2}\cos xdx

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