## Problem Statement

Calculate the definite integral:

$$
\int_{-4}^{0}\left(x^{2}+7 x+12\right) \cos x d x
$$
## Solution

$$
\int_{-4}^{0}\left(x^{2}+7 x+12\right) \cos x d x=
$$

Let:

$$
\begin{aligned}
& u=x^{2}+7 x+12 ; d u=(2 x+7) d x \\
& d v=\cos x d x ; v=\sin x
\end{aligned}
$$

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

$$
\begin{aligned}
& =\left.\left(x^{2}+7 x+12\right) \sin x\right|_{-4} ^{0}-\int_{-4}^{0} \sin x(2 x+7) d x= \\
& =\left(0^{2}+7 \cdot 0+12\right) \sin 0-\left((-4)^{2}+7 \cdot(-4)+12\right) \sin (-4)-\int_{-4}^{0} \sin x(2 x+7) d x= \\
& =12 \cdot 0-(16-28+12) \sin (-4)-\int_{-4}^{0} \sin x(2 x+7) d x=-\int_{-4}^{0} \sin x(2 x+7) d x=
\end{aligned}
$$

Let:

$$
\begin{aligned}
& u=2 x+7 ; d u=2 d x \\
& d v=\sin x d x ; v=-\cos x
\end{aligned}
$$

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

$$
=-\left(\left.(2 x+7) \cdot(-\cos x)\right|_{-4} ^{0}-\int_{-4}^{0}(-\cos x) \cdot 2 d x\right)=
$$

$$
\begin{aligned}
& =-\left((2 \cdot 0+7) \cdot(-\cos 0)-(2 \cdot(-4)+7) \cdot(-\cos (-4))+2 \int_{-4}^{0} \cos x d x\right)= \\
& =-\left(-7-\cos (-4)+\left.2 \sin x\right|_{-4} ^{0}\right)=-(-7-\cos 4+2 \sin 0-2 \sin (-4))= \\
& =-(-7-\cos 4+2 \sin 4)=7+\cos 4-2 \sin 4
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+2-5$ "

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

- Last edited: 11:59, December 9, 2009.
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## Problem Kuznetsov Integrals 2-6

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