## Problem Statement

Calculate the definite integral:

$$
\int_{\frac{\pi}{4}}^{3}\left(3 x-x^{2}\right) \sin 2 x \, d x
$$
## Solution

$$
\int_{\frac{\pi}{4}}^{3}\left(3 x-x^{2}\right) \sin 2 x d x=
$$

Let's denote:

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

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

$$
\begin{aligned}
& =\left.\left(3 x-x^{2}\right) \cdot\left(-\frac{1}{2} \cos 2 x\right)\right|_{\frac{\pi}{4}} ^{3}-\int_{\frac{\pi}{4}}^{3}\left(-\frac{1}{2} \cos 2 x\right) \cdot(3-2 x) \cdot d x= \\
& =\left(3 \cdot 3-3^{2}\right) \cdot\left(-\frac{1}{2} \cos (2 \cdot 3)\right)-\left(3 \cdot \frac{\pi}{4}-\left(\frac{\pi}{4}\right)^{2}\right) \cdot\left(-\frac{1}{2} \cos \left(2 \cdot \frac{\pi}{4}\right)\right)+\frac{1}{2} \int_{\frac{\pi}{4}}^{3} \cos 2 x \cdot(3-2 x) \cdot d x= \\
& =0 \cdot\left(-\frac{1}{2}\right) \cdot \cos 6-\left(\frac{3 \pi}{4}-\frac{\pi^{2}}{16}\right) \cdot\left(-\frac{1}{2} \cos \frac{\pi}{2}\right)+\frac{1}{2} \int_{\frac{\pi}{4}}^{3} \cos 2 x \cdot(3-2 x) \cdot d x= \\
& =-\left(\frac{3 \pi}{4}-\frac{\pi^{2}}{16}\right) \cdot\left(-\frac{1}{2}\right) \cdot 0+\frac{1}{2} \int_{\frac{\pi}{4}}^{3} \cos 2 x \cdot(3-2 x) \cdot d x=\frac{1}{2} \int_{\frac{\pi}{4}}^{3} \cos 2 x \cdot(3-2 x) \cdot d x=
\end{aligned}
$$

Let's denote:

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

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

$$
\begin{aligned}
& =\frac{1}{2} \cdot\left(\left.(3-2 x) \cdot \frac{1}{2} \sin 2 x\right|_{\frac{\pi}{4}} ^{3}-\int_{\frac{\pi}{4}}^{3} \frac{1}{2} \sin 2 x \cdot(-2) d x\right)= \\
& =\frac{1}{2} \cdot\left((3-2 \cdot 3) \cdot \frac{1}{2} \sin (2 \cdot 3)-\left(3-2 \cdot \frac{\pi}{4}\right) \cdot \frac{1}{2} \sin \left(2 \cdot \frac{\pi}{4}\right)-\int_{\frac{\pi}{4}}^{3} \sin \left(2 \cdot \frac{\pi}{4}\right) d x\right)= \\
& =\frac{1}{2} \cdot\left(-3 \cdot \frac{1}{2} \cdot \sin 6-\left(3-\frac{\pi}{2}\right) \cdot \frac{1}{2} \sin \frac{\pi}{2}-\left.\left(-\frac{1}{2} \cos 2 x\right)\right|_{\frac{\pi}{4}} ^{3}\right)= \\
& =\frac{1}{2} \cdot\left(-\frac{3}{2} \cdot \sin 6-\left(3-\frac{\pi}{2}\right) \cdot \frac{1}{2} \cdot 1+\frac{1}{2} \cos (2 \cdot 3)-\frac{1}{2} \cos \left(2 \cdot \frac{\pi}{4}\right)\right)= \\
& =\frac{1}{2} \cdot\left(-\frac{3}{2} \cdot \sin 6-\frac{3}{2}+\frac{\pi}{4}+\frac{1}{2} \cos 6-\frac{1}{2} \cos \frac{\pi}{2}\right)= \\
& =\frac{1}{2} \cdot\left(-\frac{3}{2} \cdot \sin 6-\frac{3}{2}+\frac{\pi}{4}+\frac{1}{2} \cos 6-0\right)= \\
& =-\frac{3}{4} \cdot \sin 6-\frac{3}{4}+\frac{\pi}{8}+\frac{1}{4} \cos 6=\frac{\pi-6+2 \cos 6-6 \sin 6}{8}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_2-20$ » Categories: Kuznetsov's Problem Book Integrals Problem 2 | Integrals

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- Last edited: 19:56, 27 February 2009.
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## Problem Kuznetsov Integrals 2-21

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