## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}}\left(x^{2}-5 x+6\right) \sin 3 x \, dx
$$
## Solution

$$
\int_{0}^{\frac{\pi}{2}}\left(x^{2}-5 x+6\right) \sin 3 x d x=
$$

Let:

$$
\begin{aligned}
& u=x^{2}-5 x+6 ; d u=(2 x-5) d x \\
& d v=\sin 3 x d x ; v=-\frac{1}{3} \cos 3 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =\left.\left(x^{2}-5 x+6\right) \cdot\left(-\frac{1}{3} \cos 3 x\right)\right|_{0} ^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}\left(-\frac{1}{3} \cos 3 x\right) \cdot(2 x-5) \cdot d x= \\
& =\left(\left(\frac{\pi}{2}\right)^{2}-5 \cdot \frac{\pi}{2}+6\right) \cdot\left(-\frac{1}{3} \cos \left(3 \cdot \frac{\pi}{2}\right)\right)-\left(0^{2}-5 \cdot 0+6\right) \cdot\left(-\frac{1}{3} \cos (3 \cdot 0)\right)+\frac{1}{3} \\
& =\left(\frac{\pi^{2}}{4}-\frac{5 \pi}{2}+6\right) \cdot\left(-\frac{1}{3} \cos \frac{3 \pi}{2}\right)-6 \cdot\left(-\frac{1}{3} \cos 0\right)+\frac{1}{3} \int_{0}^{\frac{\pi}{2}} \cos 3 x \cdot(2 x-5) \cdot d x= \\
& =\left(\frac{\pi^{2}}{4}-\frac{5 \pi}{2}+6\right) \cdot 0+2 \cdot 1+\frac{1}{3} \int_{0}^{\frac{\pi}{2}} \cos 3 x \cdot(2 x-5) \cdot d x=2+\frac{1}{3} \int_{0}^{\frac{\pi}{2}} \cos 3 x \cdot(2 x-5) \cdot d x=
\end{aligned}
$$

Let:

$$
\begin{aligned}
& u=2 x-5 ; d u=2 d x \\
& d v=\cos 3 x d x ; v=\frac{1}{3} \sin 3 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =2+\frac{1}{3} \cdot\left(\left.(2 x-5) \cdot \frac{1}{3} \sin 3 x\right|_{0} ^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}} \frac{1}{3} \sin 3 x \cdot 2 d x\right)= \\
& =2+\frac{1}{3} \cdot\left(\left(2 \cdot \frac{\pi}{2}-5\right) \cdot \frac{1}{3} \sin \left(3 \cdot \frac{\pi}{2}\right)-(0-5) \cdot \frac{1}{3} \sin (3 \cdot 0)-\frac{2}{3} \int_{0}^{\frac{\pi}{2}} \sin 3 x d x\right)= \\
& =2+\frac{1}{3} \cdot\left((\pi-5) \cdot \frac{1}{3} \cdot(-1)-(-5) \cdot \frac{1}{3} \cdot 0-\frac{2}{3} \int_{0}^{\frac{\pi}{2}} \sin 3 x d x\right)= \\
& =2+\frac{1}{3} \cdot\left(\frac{5-\pi}{3}-0-\left.\frac{2}{3} \cdot\left(-\frac{1}{3} \cos 3 x\right)\right|_{0} ^{\frac{\pi}{2}}\right)= \\
& =2+\frac{1}{3} \cdot\left(\frac{5-\pi}{3}+\frac{2}{9} \cos \left(3 \cdot \frac{\pi}{2}\right)-\frac{2}{9} \cos (3 \cdot 0)\right)= \\
& =2+\frac{1}{3} \cdot\left(\frac{5-\pi}{3}+\frac{2}{9} \cdot 0-\frac{2}{9} \cdot 1\right)=2+\frac{1}{3} \cdot\left(\frac{5-\pi}{3}-\frac{2}{9}\right)=2+\frac{5-\pi}{9}-\frac{2}{27}= \\
& =\frac{54}{27}+\frac{15-3 \pi}{27}-\frac{2}{27}=\frac{67-3 \pi}{27}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} \% \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}$ 2-16»

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

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- Last edited: 21:49, 26 February 2009.
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## Problem Kuznetsov Integrals 2-17

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