If int x sin x , dx = -x cos x + A,then A = | vedclass

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(A) We use the method of integration by parts for $\int x \sin x \, dx$. Let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$. Using

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$\sin x + 	ext{constant}$

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(A) We use the method of integration by parts for $\int x \sin x \, dx$. Let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$. Using the formula $\int u \, dv = uv - \int v \, du$,we get: $\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx$ $= -x \cos x + \int \cos x \, dx$ $= -x \cos x + \sin x + C$,where $C$ is the constant of integration. Comparing this with the given expression $-x \cos x + A$,we find that $A = \sin x + C$.

If $\int x \sin x \, dx = -x \cos x + A$,then $A = $

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