Evaluate the integral. int3x sin(-x)dx Answer:

Write Integral: Write down the integral to be evaluated. I = ∫3x sin(-x) dxFactor Out Constant: Use the property of integrals that ∫kf(x) dx = k∫f(x) dx, where k is a constant, to factor out the constant 3. I = 3∫x sin(-x) dxApply Trig Identity: Apply the trigonometric identity sin(-x) = -sin(x) to simplify the integrand. I = 3∫x (-sin(x)) dx I = -3∫x sin(x) dxIntegration by Parts: Use integration by parts, where u = x and dv = sin(x) dx. Then we need to find du and v. Let u = x, so du = dx. Let dv = sin(x) dx, so v = -cos(x).Find du and v: Apply the integration by parts formula ∫u dv = uv - ∫v du. I = -3(uv - ∫v du) I = -3(x(-cos(x)) - ∫(-cos(x)) dx) I = -3(-x cos(x) + ∫cos(x) dx)Apply Integration by Parts: Integrate cos(x) with respect to x. ∫cos(x) dx = sin(x)Integrate cos(x): Substitute the integral of cos(x) back into the equation. I = -3(-x cos(x) + sin(x)) + C, where C is the constant of integration.Substitute Integral: Simplify the expression. I = 3x cos(x) - 3 sin(x) + C

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Evaluate the integral.

int3x sin(-x)dx
Answer:

Evaluate the integral. $\newline$ $\int 3 x \sin (-x) d x$ $\newline$ Answer:

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Calculus

Find indefinite integrals using the substitution and by parts

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Q. Evaluate the integral.

\newline

\int 3 x \sin (-x) d x

\newline

Answer:

Write Integral: Write down the integral to be evaluated. $\newline$ $I = \int 3x \sin(-x) \, dx$
Factor Out Constant: Use the property of integrals that $\int k f(x) \, dx = k \int f(x) \, dx$ , where $k$ is a constant, to factor out the constant $3$ . $I = 3\int x \sin(-x) \, dx$
Apply Trig Identity: Apply the trigonometric identity $\sin(-x) = -\sin(x)$ to simplify the integrand. $\newline$ $I = 3\int x (-\sin(x)) \, dx$ $\newline$ $I = -3\int x \sin(x) \, dx$
Integration by Parts: Use integration by parts, where $u = x$ and $dv = \sin(x) \, dx$ . Then we need to find $du$ and $v$ . $\newline$ Let $u = x$ , so $du = dx$ . $\newline$ Let $dv = \sin(x) \, dx$ , so $v = -\cos(x)$ .
Find $du$ and $v$ : Apply the integration by parts formula $\int u\, dv = uv - \int v\, du$ .
$I = -3(uv - \int v\, du)$
$I = -3(x(-\cos(x)) - \int(-\cos(x))\, dx)$
$I = -3(-x \cos(x) + \int\cos(x)\, dx)$
Apply Integration by Parts: Integrate $\cos(x)$ with respect to $x$ . $\int\cos(x) \, dx = \sin(x)$
Integrate $\cos(x)$ : Substitute the integral of $\cos(x)$ back into the equation. $I = -3(-x \cos(x) + \sin(x)) + C$ , where $C$ is the constant of integration.
Substitute Integral: Simplify the expression. $\newline$ $I = 3x \cos(x) - 3 \sin(x) + C$