The value of the integral int x^3 cos x , dx | vedclass

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Question

(A) Using integration by parts $\int u \, dv = uv - \int v \, du$, let $u = x^3$ and $dv = \cos x \, dx$.Step $1$: $\int x^3 \cos x \, dx = x^3 \sin x

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$x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6 \cos x + c$

Vedclass · Answer

(A) Using integration by parts $\int u \, dv = uv - \int v \, du$, let $u = x^3$ and $dv = \cos x \, dx$.Step $1$: $\int x^3 \cos x \, dx = x^3 \sin x - \int 3x^2 \sin x \, dx$.Step $2$: For $\int 3x^2 \sin x \, dx$, let $u = 3x^2, dv = \sin x \, dx \implies du = 6x \, dx, v = -\cos x$.So, $\int 3x^2 \sin x \, dx = -3x^2 \cos x - \int (-6x \cos x) \, dx = -3x^2 \cos x + 6 \int x \cos x \, dx$.Step $3$: For $\int x \cos x \, dx$, let $u = x, dv = \cos x \, dx \implies du = dx, v = \sin x$.So, $\int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x$.Step $4$: Combine all parts: $x^3 \sin x - (-3x^2 \cos x + 6(x \sin x + \cos x)) + c = x^3 \sin x + 3x^2 \cos x - 6x \sin x - 6 \cos x + c$.

The value of the integral $\int x^3 \cos x \, dx$ is:

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