int x^2 cos x , dx = | vedclass

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Question

(A) To evaluate $\int x^2 \cos x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x^2$ and $dv = \cos x

Vedclass · Accepted Answer

$x^2 \sin x + 2x \cos x - 2 \sin x + c$ where $c$ is the constant of integration

Vedclass · Answer

(A) To evaluate $\int x^2 \cos x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = x^2$ and $dv = \cos x \, dx$. Then $du = 2x \, dx$ and $v = \sin x$. Applying the formula: $\int x^2 \cos x \, dx = x^2 \sin x - \int (\sin x)(2x) \, dx = x^2 \sin x - 2 \int x \sin x \, dx$. Now,apply integration by parts again for $\int x \sin x \, dx$: Let $u = x$ and $dv = \sin x \, dx$. Then $du = dx$ and $v = -\cos x$. $\int x \sin x \, dx = x(-\cos x) - \int (-\cos x) \, dx = -x \cos x + \int \cos x \, dx = -x \cos x + \sin x$. Substituting this back into the original expression: $\int x^2 \cos x \, dx = x^2 \sin x - 2(-x \cos x + \sin x) + c = x^2 \sin x + 2x \cos x - 2 \sin x + c$.

$\int x^2 \cos x \, dx =$

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