## Problem Statement

Calculate the indefinite integral:

$$
\int(4 x+3) \sin 5 x \, d x
$$
## Solution

$$
\int(4 x+3) \sin 5 x d x=
$$

Let's denote:

$$
\begin{aligned}
& u=4 x+3 ; d u=4 d x \\
& d v=\sin 5 x d x ; v=-\frac{1}{5} \cos 5 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =(4 x+3) \cdot\left(-\frac{1}{5} \cos 5 x\right)-\int\left(-\frac{1}{5} \cos 5 x\right) \cdot 4 d x= \\
& =-\frac{1}{5} \cdot(4 x+3) \cos 5 x+\frac{4}{5} \int \cos 5 x d x=-\frac{1}{5} \cdot(4 x+3) \cos 5 x+\frac{4}{5} \cdot \frac{1}{5} \cdot \sin 5 x+C= \\
& =-\frac{1}{5} \cdot(4 x+3) \cos 5 x+\frac{4}{25} \cdot \sin 5 x+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-25$ » Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals

Ukrainian Banner Network

- Last edited on this page: 14:50, 20 February 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 1-26

## Material from Plusi