## Problem Statement

Calculate the indefinite integral:

$$
\int(\sqrt{2}-8 x) \sin 3 x \, d x
$$
## Solution

$$
\int(\sqrt{2}-8 x) \sin 3 x d x=
$$

Let:

$$
\begin{aligned}
& u=\sqrt{2}-8 x ; d u=-8 d x \\
& d v=\sin 3 x d x ; v=-\frac{1}{3} \cos 3 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =(\sqrt{2}-8 x) \cdot\left(-\frac{1}{3} \cos 3 x\right)-\int\left(-\frac{1}{3} \cos 3 x\right) \cdot(-8) d x= \\
& =\frac{1}{3} \cdot(8 x-\sqrt{2}) \cos 3 x-\frac{8}{3} \int \cos 3 x d x=\frac{1}{3} \cdot(8 x-\sqrt{2}) \cos 3 x-\frac{8}{3} \cdot \frac{1}{3} \cdot \sin 3 x+C= \\
& =\frac{1}{3} \cdot(8 x-\sqrt{2}) \cos 3 x-\frac{8}{9} \cdot \sin 3 x+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-27$ »

Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals | Problems for Checking

- Last modified: 08:56, 9 July 2009.
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## Problem Kuznetsov Integrals $1-28$

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