## Condition of the problem

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{4}}\left(x^{2}+17.5\right) \sin 2 x d x
$$
## Solution

$$
\int_{0}^{\frac{\pi}{4}}\left(x^{2}+17.5\right) \sin 2 x d x=
$$

Let's denote:

$$
\begin{aligned}
& u=x^{2}+17.5 ; d u=2 x \cdot d x \\
& d v=\sin 2 x d x ; v=-\frac{1}{2} \cos 2 x
\end{aligned}
$$

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

$$
\begin{aligned}
& =\left.\left(x^{2}+17.5\right) \cdot\left(-\frac{1}{2} \cos 2 x\right)\right|_{0} ^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\left(-\frac{1}{2} \cos 2 x\right) \cdot 2 x \cdot d x= \\
& =\left(\left(\frac{\pi}{4}\right)^{2}+17.5\right) \cdot\left(-\frac{1}{2} \cos \left(2 \cdot \frac{\pi}{4}\right)\right)-\left(0^{2}+17.5\right) \cdot\left(-\frac{1}{2} \cos (2 \cdot 0)\right)+\int_{0}^{\frac{\pi}{4}} \cos 2 x \cdot x \cdot d x= \\
& =\left(\frac{\pi^{2}}{16}+17.5\right) \cdot\left(-\frac{1}{2} \cos \frac{\pi}{2}\right)-17.5 \cdot\left(-\frac{1}{2} \cos 0\right)+\int_{0}^{\frac{\pi}{4}} \cos 2 x \cdot x \cdot d x= \\
& =\left(\frac{\pi^{2}}{16}+17.5\right) \cdot\left(-\frac{1}{2}\right) \cdot 0+17.5 \cdot \frac{1}{2} \cdot 1+\int_{0}^{\frac{\pi}{4}} \cos 2 x \cdot x \cdot d x= \\
& =\frac{35}{4}+\int_{0}^{\frac{\pi}{4}} \cos 2 x \cdot x \cdot d x=
\end{aligned}
$$

Let's denote:

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

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

$$
\begin{aligned}
& =\frac{35}{4}+\left.\left(x \cdot \frac{1}{2} \sin 2 x\right)\right|_{0} ^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}} \frac{1}{2} \sin 2 x \cdot d x= \\
& =\frac{35}{4}+\frac{\pi}{4} \cdot \frac{1}{2} \sin \left(2 \cdot \frac{\pi}{4}\right)-0 \cdot \frac{1}{2} \sin (2 \cdot 0)-\frac{1}{2} \cdot \int_{0}^{\frac{\pi}{4}} \sin 2 x \cdot d x= \\
& =\frac{35}{4}+\frac{\pi}{4} \cdot \frac{1}{2} \sin \frac{\pi}{2}-0-\left.\frac{1}{2} \cdot\left(-\frac{1}{2} \cdot \cos 2 x\right)\right|_{0} ^{\frac{\pi}{4}}= \\
& =\frac{35}{4}+\frac{\pi}{4} \cdot \frac{1}{2} \cdot 1+\frac{1}{4} \cdot \cos \left(2 \cdot \frac{\pi}{4}\right)-\frac{1}{4} \cdot \cos (2 \cdot 0)= \\
& =\frac{35}{4}+\frac{\pi}{8}+\frac{1}{4} \cdot \cos \frac{\pi}{2}-\frac{1}{4} \cdot \cos 0=\frac{35}{4}+\frac{\pi}{8}+\frac{1}{4} \cdot 0-\frac{1}{4} \cdot 1= \\
& =\frac{35}{4}+\frac{\pi}{8}-\frac{1}{4}=\frac{34}{4}+\frac{\pi}{8}=\frac{17}{2}+\frac{\pi}{8}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}$ 2-18» Categories: Kuznetsov's Problem Book Integrals Problem $2 \mid$ Integrals

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- Last edited: 19:24, 27 February 2009.
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## Problem Kuznetsov Integrals 2-19

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