Use integration by parts to evaluate: ∫xsin(x) dx.

Since our function is a product of two "mini-functions" of x, we are able to use integration by parts.The trick for this is to correctly set 'u' and 'dv'. 'u' should be labelled as the function which can reduce when differentiated. This means that the function should decrease in power. From our main function, we have both x, and sin(x). If we differentiate sin(x), we get cos(x), which hasn't decreased. However, if we differentiate x, we get 1, which has decreased in power.Using the integration by parts formula: ∫[udv] = uv - ∫[vdu].By setting u=x, and dv=sin(x), we can calculate 'du' and 'v' by differentiating u, and integrating dv respectively.u=x --> du=1, and dv=sin(x) --> v= -cos(x).Substituting this into the integration by parts formula stated above gives:∫[xsin(x)] = x*(-cos(x)) - ∫[(-cos(x)1)],Therefore ∫[xsin(x)] = -xcos(x) - ∫[(-cos(x)],And finally: ∫[xsin(x)] = -xcos(x) - (-sin(x)).This gives us our final answer of:∫[xsin(x)] = -xcos(x)+sin(x) + C, where C is the constant of integration.

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Answered by Bailey A. Maths tutor

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