## Problem Statement

Calculate the definite integral:

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

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

Let's denote:

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

Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:

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

Let's denote:

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

Using the integration by parts formula \(\int u d v=u v-\int v d u\), we get:

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

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

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} 2-14 »

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

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## Problem Kuznetsov Integrals 2-15

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