Evaluate the integral. int-6x sin(-x)dx Answer:

Rewrite Integral: Rewrite the integral by taking the constant -6 outside and using the property sin(-x) = -sin(x). ∫-6x sin(-x)dx = -6 ∫x (-sin(x))dx = 6 ∫x sin(x)dxIntegration by Parts: Apply integration by parts, where u = x and dv = sin(x)dx. Then, du = dx and v = -cos(x). Using the integration by parts formula ∫udv = uv - ∫vdu, we get: ∫x sin(x)dx = x(-cos(x)) - ∫(-cos(x))(dx) = -x cos(x) + ∫cos(x)dxIntegrate Cos: Integrate cos(x) with respect to x. ∫cos(x)dx = sin(x)Substitute Integral: Substitute the integral of cos(x) back into the equation. -6 ( -x cos(x) + sin(x) ) + C = 6x cos(x) - 6 sin(x) + C, where C is the constant of integration.Final Answer: Write the final answer. The integral of -6x sin(-x) with respect to x is 6x cos(x) - 6 sin(x) + C.

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Calculus

Find indefinite integrals using the substitution and by parts

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Q. Evaluate the integral.

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\int-6 x \sin (-x) d x

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Answer:

Rewrite Integral: Rewrite the integral by taking the constant $-6$ outside and using the property $\sin(-x) = -\sin(x)$ . $\int -6x \sin(-x)\,dx = -6 \int x (-\sin(x))\,dx$ $= 6 \int x \sin(x)\,dx$
Integration by Parts: Apply integration by parts, where $u = x$ and $dv = \sin(x)dx$ . Then, $du = dx$ and $v = -\cos(x)$ . Using the integration by parts 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
Integrate Cos: Integrate \(\cos(x) with respect to $x$ . $\int\cos(x)\,dx = \sin(x)$
Substitute Integral: Substitute the integral of $\cos(x)$ back into the equation. $-6 ( -x \cos(x) + \sin(x) ) + C = 6x \cos(x) - 6 \sin(x) + C$ , where $C$ is the constant of integration.
Final Answer: Write the final answer. $\newline$ The integral of $-6x \sin(-x)$ with respect to $x$ is $6x \cos(x) - 6 \sin(x) + C$ .