Evaluate the integral. int-x cos(-3x)dx Answer:

Write Integral: Write down the integral to be solved. I = ∫-x cos(-3x) 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 -1 from the integral. I = -∫x cos(-3x) dxApply Even Property: Recognize that cos(-θ) = cos(θ) due to the even property of the cosine function. I = -∫x cos(3x) dxApply Integration by Parts: Apply integration by parts, where u = x and dv = cos(3x) dx. Then we need to find du and v. Let u = x, so du = dx. Let dv = cos(3x) dx, so v = (1/3)sin(3x), since the integral of cos(3x) dx is (1/3)sin(3x).Integrate sin(3x): Apply the integration by parts formula: ∫u dv = uv - ∫v du. I = -(u * v - ∫v du) I = -(x * (1/3)sin(3x) - ∫(1/3)sin(3x) dx)Simplify and Combine: Integrate (1/3)sin(3x) with respect to x. The integral of sin(3x) dx is -(1/3)cos(3x), so the integral of (1/3)sin(3x) dx is -(1/9)cos(3x). I = -(x * (1/3)sin(3x) - -(1/9)cos(3x))Factor Out Constants: Simplify the expression and combine like terms. I = -(1/3)x sin(3x) + (1/9)cos(3x) + C, where C is the constant of integration.Factor out the constants to make the expression cleaner. I = -1/3 x sin(3x) + 1/9 cos(3x) + C

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Evaluate the integral.

int-x cos(-3x)dx
Answer:

Evaluate the integral. $\newline$ $\int-x \cos (-3 x) d x$ $\newline$ 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-x \cos (-3 x) d x

\newline

Answer:

Write Integral: Write down the integral to be solved. $\newline$ $I = \int -x \cos(-3x) \, dx$
Factor Out Constant: Use the property of integrals that $\int kf(x) \, dx = k\int f(x) \, dx$ , where $k$ is a constant, to factor out the constant $-1$ from the integral. $\newline$ $I = -\int x \cos(-3x) \, dx$
Apply Even Property: Recognize that $\cos(-\theta) = \cos(\theta)$ due to the even property of the cosine function. $\newline$ $I = -\int x \cos(3x) \, dx$
Apply Integration by Parts: Apply integration by parts, where $u = x$ and $dv = \cos(3x) dx$ . Then we need to find $du$ and $v$ . $\newline$ Let $u = x$ , so $du = dx$ . $\newline$ Let $dv = \cos(3x) dx$ , so $v = \frac{1}{3}\sin(3x)$ , since the integral of $\cos(3x) dx$ is $\frac{1}{3}\sin(3x)$ .
Integrate $\sin(3x)$ : Apply the integration by parts formula: $\int u \, dv = uv - \int v \, du$ . $\newline$ $I = -(u \cdot v - \int v \, du)$ $\newline$ I = -\left(x \cdot \left(\frac{$1$}{$3$}\right)\sin(\(3x) - \int\left(\frac{ $1$ }{ $3$ }\right)\sin( $3$ x) \, dx\right)
Simplify and Combine: Integrate $(\frac{1}{3})\sin(3x)$ with respect to $x$ . The integral of $\sin(3x)$ $dx$ is $-\frac{1}{3}\cos(3x)$ , so the integral of $(\frac{1}{3})\sin(3x)$ $dx$ is $-\frac{1}{9}\cos(3x)$ . $I = -\left(x \cdot (\frac{1}{3})\sin(3x) - -\frac{1}{9}\cos(3x)\right)$
Factor Out Constants: Simplify the expression and combine like terms. $\newline$ $I = -(\frac{1}{3})x \sin(3x) + (\frac{1}{9})\cos(3x) + C$ , where $C$ is the constant of integration.
Factor Out Constants: Simplify the expression and combine like terms. $\newline$ $I = -(\frac{1}{3})x \sin(3x) + (\frac{1}{9})\cos(3x) + C$ , where $C$ is the constant of integration.Factor out the constants to make the expression cleaner. $\newline$ $I = -\frac{1}{3} x \sin(3x) + \frac{1}{9} \cos(3x) + C$