 # What is &#x222B;<!-- ∫ --> measgachyx5q9 2022-05-06 Answered
What is $\int \frac{8}{9}﻿x\mathrm{cos}\left(3x\right)dx$?
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Simplify.

$\int \frac{8x\mathrm{cos}\left(3x\right)}{9}dx$

Since $\frac{8}{9}$ is constant with respect to $x$, move $\frac{8}{9}$ out of the integral.

$\frac{8}{9}\int x\mathrm{cos}\left(3x\right)dx$

Integrate by parts using the formula $\int udv=uv-\int vdu$, where $u=x$ and $dv=\mathrm{cos}\left(3x\right)$.

$\frac{8}{9}\left(x\left(\frac{1}{3}\mathrm{sin}\left(3x\right)\right)-\int \frac{1}{3}\mathrm{sin}\left(3x\right)dx\right)$

Simplify.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\int \frac{\mathrm{sin}\left(3x\right)}{3}dx\right)$

Since $\frac{1}{3}$ is constant with respect to $x$, move $\frac{1}{3}$ out of the integral.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\left(\frac{1}{3}\int \mathrm{sin}\left(3x\right)dx\right)\right)$

Let $u=3x$. Then $du=3dx$, so $\frac{1}{3}du=dx$. Rewrite using $u$ and $d$$u$.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\frac{1}{3}\int \mathrm{sin}\left(u\right)\frac{1}{3}du\right)$

Combine $\mathrm{sin}\left(u\right)$ and $\frac{1}{3}$.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\frac{1}{3}\int \frac{\mathrm{sin}\left(u\right)}{3}du\right)$

Since $\frac{1}{3}$ is constant with respect to $u$, move $\frac{1}{3}$ out of the integral.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\frac{1}{3}\left(\frac{1}{3}\int \mathrm{sin}\left(u\right)du\right)\right)$

Simplify.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\frac{1}{9}\int \mathrm{sin}\left(u\right)du\right)$

The integral of $\mathrm{sin}\left(u\right)$ with respect to $u$ is $-\mathrm{cos}\left(u\right)$.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\frac{1}{9}\left(-\mathrm{cos}\left(u\right)+C\right)\right)$

Rewrite $\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}-\frac{1}{9}\left(-\mathrm{cos}\left(u\right)+C\right)\right)$ as $\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}+\frac{\mathrm{cos}\left(u\right)}{9}\right)+C$.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}+\frac{\mathrm{cos}\left(u\right)}{9}\right)+C$

Replace all occurrences of $u$ with $3x$.

$\frac{8}{9}\left(\frac{x\mathrm{sin}\left(3x\right)}{3}+\frac{\mathrm{cos}\left(3x\right)}{9}\right)+C$

Simplify.

$\frac{8x\mathrm{sin}\left(3x\right)}{27}+\frac{8\mathrm{cos}\left(3x\right)}{81}+C$

Reorder terms.

$\frac{8}{27}x\mathrm{sin}\left(3x\right)+\frac{8}{81}\mathrm{cos}\left(3x\right)+C$