## Problem Statement

Calculate the definite integral:

$$
\int_{-2}^{0}(x+2)^{2} \cos 3 x \, d x
$$
## Solution

$$
\int_{-2}^{0}(x+2)^{2} \cos 3 x d x=
$$

Let's denote:

$$
\begin{aligned}
& u=(x+2)^{2} ; d u=2(x+2) \cdot d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =\left.(x+2)^{2} \cdot \frac{1}{3} \sin 3 x\right|_{-2} ^{0}-\int_{-2}^{0} \frac{1}{3} \sin 3 x \cdot 2(x+2) \cdot d x= \\
& =(0+2)^{2} \cdot \frac{1}{3} \sin (3 \cdot 0)-(-2+2)^{2} \cdot \frac{1}{3} \sin (3 \cdot(-2))-\frac{2}{3} \int_{-2}^{0} \sin 3 x \cdot(x+2) \cdot d x= \\
& =4 \cdot \frac{1}{3} \cdot 0-0 \cdot \frac{1}{3} \sin (-6)-\frac{2}{3} \int_{-2}^{0} \sin 3 x \cdot(x+2) \cdot d x=-\frac{2}{3} \int_{-2}^{0} \sin 3 x \cdot(x+2) \cdot d x=
\end{aligned}
$$

Let's denote:

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

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

$$
\begin{aligned}
& =-\frac{2}{3} \cdot\left(\left.(x+2) \cdot\left(-\frac{1}{3} \cos 3 x\right)\right|_{-2} ^{0}-\int_{-2}^{0}\left(-\frac{1}{3} \cos 3 x\right) d x\right)= \\
& =-\frac{2}{3} \cdot\left((0+2) \cdot\left(-\frac{1}{3} \cos (3 \cdot 0)\right)-(-2+2) \cdot\left(-\frac{1}{3} \cos (3 \cdot(-2))\right)+\frac{1}{3} \int_{-2}^{0} \cos 3 x d x\right)=
\end{aligned}
$$

$$
\begin{aligned}
& =-\frac{2}{3} \cdot\left(-\frac{2}{3}-0+\left.\frac{1}{3} \cdot \frac{1}{3} \sin 3 x\right|_{-2} ^{0}\right)= \\
& =-\frac{2}{3} \cdot\left(-\frac{2}{3}+\frac{1}{9} \sin (3 \cdot 0)-\frac{1}{9} \sin (3 \cdot(-2))\right)= \\
& =-\frac{2}{3} \cdot\left(-\frac{2}{3}+\frac{1}{9} \sin 6\right)=\frac{4}{9}-\frac{2}{27} \sin 6=\frac{12-2 \sin 6}{27}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 2-4$ »

Categories: Kuznetsov's Problem Book Integrals Problem 2 | Integrals

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- Last edited: 19:47, May 5, 2009.
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## Problem Kuznetsov Integrals $2-5$

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