## Problem Statement

Calculate the definite integral:

$$
\int_{-2}^{0}\left(x^{2}+5 x+6\right) \cos 2 x \, d x
$$
## Solution

$$
\int_{-2}^{0}\left(x^{2}+5 x+6\right) \cos 2 x d x=
$$

Let's denote:

$$
\begin{aligned}
& u=x^{2}+5 x+6 ; d u=(2 x+5) d x \\
& d v=\cos 2 x d x ; v=\frac{1}{2} \sin 2 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =\left.\left(x^{2}+5 x+6\right) \cdot \frac{1}{2} \sin 2 x\right|_{-2} ^{0}-\int_{-2}^{0} \frac{1}{2} \sin 2 x(2 x+5) d x= \\
& =\left(0^{2}+5 \cdot 0+6\right) \cdot \frac{1}{2} \sin (2 \cdot 0)-\left((-2)^{2}+5 \cdot(-2)+6\right) \cdot \frac{1}{2} \sin (2 \cdot(-2))-\frac{1}{2} \int_{-2}^{0} \sin 2 x(2 x+5) d x= \\
& =6 \cdot \frac{1}{2} \cdot 0-(4-10+6) \cdot \frac{1}{2} \sin (-4)-\frac{1}{2} \int_{-2}^{0} \sin 2 x(2 x+5) d x=-\frac{1}{2} \int_{-2}^{0} \sin 2 x(2 x+5) d x=
\end{aligned}
$$

Let's denote:

$$
\begin{aligned}
& u=2 x+5 ; d u=2 d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =-\frac{1}{2} \cdot\left(\left.(2 x+5) \cdot\left(-\frac{1}{2} \cos 2 x\right)\right|_{-2} ^{0}-\int_{-2}^{0}\left(-\frac{1}{2} \cos 2 x\right) \cdot 2 d x\right)= \\
& =-\frac{1}{2} \cdot\left((2 \cdot 0+5) \cdot\left(-\frac{1}{2} \cos (2 \cdot 0)\right)-(2 \cdot(-2)+5) \cdot\left(-\frac{1}{2} \cos (2 \cdot(-2))\right)+\int_{-2}^{0} \cos 2 x d x\right)= \\
& =-\frac{1}{2} \cdot\left(-\frac{5}{2}+\frac{\cos (-4)}{2}+\left.\frac{1}{2} \sin 2 x\right|_{-2} ^{0}\right)=-\frac{1}{2} \cdot\left(-\frac{5}{2}+\frac{\cos (4)}{2}+\frac{1}{2} \sin (2 \cdot 0)-\frac{1}{2} \sin (2 \cdot(-2))\right)= \\
& =-\frac{1}{2} \cdot\left(-\frac{5}{2}+\frac{\cos 4}{2}+\frac{1}{2} \sin 4\right)=\frac{5-\cos 4-\sin 4}{4}
\end{aligned}
$$

$\% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_2-1 \%$

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

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## Problem Kuznetsov Integrals 2-2

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