Evaluate the integral: int (e^{log(sin x)} + | vedclass

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(B) Given integral is $I = \int (e^{\log(\sin x)} + \cos x) x \, dx$.Since $e^{\log(\sin x)} = \sin x$, the integral becomes $I = \int (\sin x + \cos

Vedclass · Accepted Answer

$x(\sin x - \cos x) + (\sin x + \cos x) + c$

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(B) Given integral is $I = \int (e^{\log(\sin x)} + \cos x) x \, dx$.Since $e^{\log(\sin x)} = \sin x$, the integral becomes $I = \int (\sin x + \cos x) x \, dx$.$I = \int x \sin x \, dx + \int x \cos x \, dx$.Using integration by parts $\int u v \, dx = u \int v \, dx - \int (u' \int v \, dx) \, dx$:For $\int x \sin x \, dx = x(-\cos x) - \int (1)(-\cos x) \, dx = -x \cos x + \sin x$.For $\int x \cos x \, dx = x(\sin x) - \int (1)(\sin x) \, dx = x \sin x + \cos x$.Adding these results: $I = (-x \cos x + \sin x) + (x \sin x + \cos x) + c$.$I = x(\sin x - \cos x) + (\sin x + \cos x) + c$.

Evaluate the integral: $\int (e^{\log(\sin x)} + \cos x) x \, dx$

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