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Evaluate the integral. int-4x sin(4x+5)dx Answer:

Evaluate the integral.\newline∫−4xsin⁡(4x+5)dx \int-4 x \sin (4 x+5) d x \newlineAnswer:

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Full solution

Q. Evaluate the integral.\newline∫−4xsin⁡(4x+5)dx \int-4 x \sin (4 x+5) d x \newlineAnswer:
  1. Choose uu and dvdv: Let's use integration by parts to solve the integral of the form ∫udv\int u dv. We can choose u=−4xu = -4x and dv=sin⁡(4x+5)dxdv = \sin(4x+5)dx. Then we need to find dudu and vv.
    u=−4xu = -4x implies du=−4dxdu = -4 dx
    To find vv, we integrate dvdv:
    dvdv11
    To integrate dvdv22, we use a substitution. Let dvdv33, then dvdv44, or dvdv55.
    dvdv66
    dvdv77
    Now we substitute back dvdv33:
    dvdv99
  2. Integrate dv to find v: Now we apply the integration by parts formula:\newline∫udv=uv−∫vdu\int u dv = uv - \int v du\newlinePlugging in our uu, vv, dudu, and dvdv, we get:\newline∫(−4x)sin⁡(4x+5)dx=−4x⋅(−14cos⁡(4x+5))−∫(−14cos⁡(4x+5))⋅(−4)dx\int (-4x) \sin(4x+5)dx = -4x \cdot \left(-\frac{1}{4}\cos(4x+5)\right) - \int \left(-\frac{1}{4}\cos(4x+5)\right) \cdot (-4)dx\newlineSimplify the equation:\newline∫(−4x)sin⁡(4x+5)dx=xcos⁡(4x+5)−∫cos⁡(4x+5)dx\int (-4x) \sin(4x+5)dx = x\cos(4x+5) - \int \cos(4x+5)dx
  3. Apply integration by parts formula: Now we need to integrate cos⁡(4x+5)\cos(4x+5). We will use the same substitution as before:\newlinew=4x+5w = 4x+5, dw=4 dxdw = 4\,dx, or dx=dw4dx = \frac{dw}{4}.\newline∫cos⁡(4x+5) dx=∫cos⁡(w)⋅(14)dw\int \cos(4x+5)\,dx = \int \cos(w) \cdot \left(\frac{1}{4}\right)dw\newlineThis integral is straightforward:\newline∫cos⁡(w)⋅(14)dw=(14)sin⁡(w)\int \cos(w) \cdot \left(\frac{1}{4}\right)dw = \left(\frac{1}{4}\right)\sin(w)\newlineSubstitute back w=4x+5w = 4x+5:\newline∫cos⁡(4x+5) dx=(14)sin⁡(4x+5)\int \cos(4x+5)\,dx = \left(\frac{1}{4}\right)\sin(4x+5)
  4. Integrate cos⁡(4x+5)\cos(4x+5): Now we substitute the integral we just found back into our integration by parts formula:\newline∫(−4x)sin⁡(4x+5) dx=xcos⁡(4x+5)−(14)sin⁡(4x+5)\int(-4x) \sin(4x+5)\,dx = x\cos(4x+5) - (\frac{1}{4})\sin(4x+5)\newlineThis is the antiderivative of the function. We add the constant of integration CC to complete the indefinite integral.

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