The value of int x sin kx , dx is | vedclass

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Question

(D) To evaluate the integral $I = \int x \sin kx \, dx$,we use the method of integration by parts,which is given by $\int u \, dv = uv - \int v \, du$

Vedclass · Accepted Answer

$-\frac{x \cos kx}{k} + \frac{\sin kx}{k^2} + c$

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(D) To evaluate the integral $I = \int x \sin kx \, dx$,we use the method of integration by parts,which is given by $\int u \, dv = uv - \int v \, du$.Let $u = x$ and $dv = \sin kx \, dx$.Then,$du = dx$ and $v = \int \sin kx \, dx = -\frac{\cos kx}{k}$.Applying the formula:$I = x \left( -\frac{\cos kx}{k} \right) - \int \left( -\frac{\cos kx}{k} \right) dx$$I = -\frac{x \cos kx}{k} + \frac{1}{k} \int \cos kx \, dx$$I = -\frac{x \cos kx}{k} + \frac{1}{k} \left( \frac{\sin kx}{k} \right) + c$$I = -\frac{x \cos kx}{k} + \frac{\sin kx}{k^2} + c$.

The value of $\int x \sin kx \, dx$ is

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