Evaluate the integral. int-2x cos(-2x)dx Answer:

Simplify Integral: Let's first simplify the integral by using the property of cosine that cos(-θ) = cos(θ). This simplifies our integral to: ∫-2x cos(-2x)dx = ∫-2x cos(2x)dx Now we can proceed with integration by parts, where we let u = -2x and dv = cos(2x)dx. Then we need to find du and v. Differentiating u with respect to x gives us du = -2dx, and integrating dv gives us v = (1/2)sin(2x), since the integral of cos(2x)dx is (1/2)sin(2x).Integration by Parts: Now we apply the integration by parts formula: ∫udv = uv - ∫vdu Substituting our chosen u, v, du, and dv, we get: ∫-2x cos(2x)dx = (-2x)(1/2)sin(2x) - ∫(1/2)sin(2x)(-2dx) Simplifying the right side, we have: -2x * (1/2)sin(2x) + ∫sin(2x)dxIntegrate sin(2x): Next, we integrate ∫sin(2x)dx. The integral of sin(2x) with respect to x is (-1/2)cos(2x), so we have: -2x * (1/2)sin(2x) + (-1/2)cos(2x) + C where C is the constant of integration.Combine Terms: Now we combine the terms to get the final answer: -2x * (1/2)sin(2x) + (-1/2)cos(2x) + C simplifies to -xsin(2x) - (1/2)cos(2x) + C This is our final answer.

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Evaluate the integral.

int-2x cos(-2x)dx
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Evaluate the integral. $\newline$ $\int-2 x \cos (-2 x) d x$ $\newline$ 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-2 x \cos (-2 x) d x

\newline

Answer:

Simplify Integral: Let's first simplify the integral by using the property of cosine that $\cos(-\theta) = \cos(\theta)$ . This simplifies our integral to: $\newline$ $\int -2x \cos(-2x)\,dx = \int -2x \cos(2x)\,dx$ $\newline$ Now we can proceed with integration by parts, where we let $u = -2x$ and $dv = \cos(2x)\,dx$ . Then we need to find $du$ and $v$ . $\newline$ Differentiating $u$ with respect to $x$ gives us $du = -2\,dx$ , and integrating $dv$ gives us $\int -2x \cos(-2x)\,dx = \int -2x \cos(2x)\,dx$ $0$ , since the integral of $\int -2x \cos(-2x)\,dx = \int -2x \cos(2x)\,dx$ $1$ is $\int -2x \cos(-2x)\,dx = \int -2x \cos(2x)\,dx$ $2$ .
Integration by Parts: Now we apply the integration by parts formula: $\newline$ $\int u\,dv = uv - \int v\,du$ $\newline$ Substituting our chosen $u$ , $v$ , $du$ , and $dv$ , we get: $\newline$ $\int -2x \cos(2x)\,dx = (-2x)(\frac{1}{2})\sin(2x) - \int (\frac{1}{2})\sin(2x)(-2\,dx)$ $\newline$ Simplifying the right side, we have: $\newline$ $-2x \times (\frac{1}{2})\sin(2x) + \int \sin(2x)\,dx$
Integrate $\sin(2x)$ : Next, we integrate $\int \sin(2x) \, dx$ . The integral of $\sin(2x)$ with respect to $x$ is $(-\frac{1}{2})\cos(2x)$ , so we have: $\newline$ $-2x \cdot (\frac{1}{2})\sin(2x) + (-\frac{1}{2})\cos(2x) + C$ $\newline$ where $C$ is the constant of integration.
Combine Terms: Now we combine the terms to get the final answer: $\newline$ $-2x \times (\frac{1}{2})\sin(2x) + (\frac{-1}{2})\cos(2x) + C$ simplifies to $\newline$ $-x\sin(2x) - (\frac{1}{2})\cos(2x) + C$ $\newline$ This is our final answer.