## Problem Statement

Calculate the definite integral:

$$
\int_{-1}^{0}\left(x^{2}+2 x+1\right) \sin 3 x d x
$$
## Solution

$$
\int_{-1}^{0}\left(x^{2}+2 x+1\right) \sin 3 x d x=
$$

Let's denote:

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

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

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

Let's denote:

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

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

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

$$
\begin{aligned}
& =-\frac{1}{3}+\frac{2}{3} \cdot\left((0+1) \cdot \frac{1}{3} \sin (3 \cdot 0)-(-1+1) \cdot \frac{1}{3} \sin (3 \cdot(-1))-\frac{1}{3} \int_{-1}^{0} \sin 3 x d x\right)= \\
& =-\frac{1}{3}+\frac{2}{3} \cdot\left(1 \cdot \frac{1}{3} \cdot 0-0 \cdot \frac{1}{3} \sin (-3)-\left.\frac{1}{3} \cdot\left(-\frac{1}{3} \cos 3 x\right)\right|_{-1} ^{0}\right)= \\
& =-\frac{1}{3}+\frac{2}{3} \cdot\left(0-0+\frac{1}{9} \cos (3 \cdot 0)-\frac{1}{9} \cos (3 \cdot(-1))\right)= \\
& =-\frac{1}{3}+\frac{2}{3} \cdot\left(\frac{1}{9}-\frac{1}{9} \cos (-3)\right)=-\frac{1}{3}+\frac{2}{27}-\frac{2}{27} \cos 3=\frac{-7-2 \cos 3}{27}=-\frac{7+2 \cos 3}{27}
\end{aligned}
$$

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%D1%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 2-13$ " Categories: Kuznetsov's Problem Book Integrals Problem 2 | Integrals

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## Problem Kuznetsov Integrals 2-14

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