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int4x cos(2-3x)dx

$\int 4 x \cos (2-3 x) d x$

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Find indefinite integrals using the substitution and by parts

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Q.

\int 4 x \cos (2-3 x) d x

Set up integral: Let's start by setting up the integral we need to solve: $\int 4x \cos(2-3x)\,dx$ . We'll use a substitution method here. Let $u = 2 - 3x$ , then $du = -3 \,dx$ , or $dx = -\frac{du}{3}$ .
Substitute $u$ and $dx$ : Substitute $u$ and $dx$ in the integral: $\int(4x \cos(u) * (-\frac{1}{3}) \,du)$ . This simplifies to $-\frac{4}{3} \int(x \cos(u) \,du)$ . Now, we need to express $x$ in terms of $u$ . From $u = 2 - 3x$ , we solve for $x$ : $dx$ $0$ .
Express $x$ in terms of $u$ : Plug $x$ back into the integral: $-\frac{4}{3} \int\left(\frac{2-u}{3} \cos(u) \,du\right)$ . Simplify the constants: $-\frac{4}{9} \int((2-u) \cos(u) \,du)$ . Now, we need to distribute the $\cos(u)$ inside the integral: $-\frac{4}{9} \left(\int(2\cos(u) \,du) - \int(u \cos(u) \,du)\right)$ .
Distribute $\cos(u)$ inside integral: Solve the first integral $\int(2\cos(u) \, du)$ . This is straightforward: $2\sin(u)$ . For the second integral, $\int(u \cos(u) \, du)$ , use integration by parts. Let $v = u$ , $dv = \cos(u) \, du$ , then $du = dv$ , and $v = \sin(u)$ .

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