## Problem Statement

Calculate the indefinite integral:

$$
\int(4 x+7) \cos 3 x \, d x
$$
## Solution

$$
\int(4 x+7) \cos 3 x d x=
$$

Let's denote:

$$
\begin{aligned}
& u=4 x+7 ; d u=4 d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$

We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:

$$
\begin{aligned}
& =(4 x+7) \cdot \frac{1}{3} \sin 3 x-\int \frac{1}{3} \sin 3 x \cdot 4 d x=\frac{1}{3} \cdot(4 x+7) \sin 3 x-\frac{4}{3} \int \sin 3 x d x= \\
& =\frac{1}{3} \cdot(4 x+7) \sin 3 x+\frac{4}{3} \cdot \frac{1}{3} \cdot \cos 3 x+C=\frac{1}{3} \cdot(4 x+7) \sin 3 x+\frac{4}{9} \cdot \cos 3 x+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-20$ » Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals

- Last edited: 12:42, 20 February 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 1-21

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