## Problem Statement

Calculate the definite integral:

$$
\int_{-3}^{3} x^{2} \cdot \sqrt{9-x^{2}} d x
$$
## Solution

$$
\int_{-3}^{3} x^{2} \cdot \sqrt{9-x^{2}} d x=
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_543e00be326595f7ce31g-34.jpg?height=1468&width=994&top_left_y=1032&top_left_x=949)

Integrals 12-20

Substitution:

$$
x=3 \sin t \Rightarrow d x=3 \cos t d t
$$

$$
\begin{aligned}
& x=-3 \Rightarrow t=\arcsin \frac{-3}{3}=-\frac{\pi}{2} \\
& x=3 \Rightarrow t=\arcsin \frac{3}{3}=\frac{\pi}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{-\pi / 2}^{\pi / 2} 9 \sin ^{2} t \cdot \sqrt{9-9 \sin ^{2} t} \cdot 3 \cos t d t=\int_{-\pi / 2}^{\pi / 2} 9 \sin ^{2} t \cdot 3 \cos t \cdot 3 \cos t d t= \\
& =81 \cdot \int_{-\pi / 2}^{\pi / 2} \sin ^{2} t \cdot \cos ^{2} t d t=\frac{81}{4} \cdot \int_{-\pi / 2}^{\pi / 2} \sin ^{2} 2 t d t=\frac{81}{4} \cdot \int_{-\pi / 2}^{\pi / 2} \frac{1-\cos 4 t}{2} d t= \\
& =\frac{81}{8} \cdot \int_{-\pi / 2}^{\pi / 2}(1-\cos 4 t) d t=\left.\frac{81}{8}\left(t-\frac{1}{4} \sin 4 t\right)\right|_{-\pi / 2} ^{\pi / 2}= \\
& =\frac{81}{8}\left(\frac{\pi}{2}-\frac{1}{4} \sin 2 \pi\right)-\frac{81}{8}\left(-\frac{\pi}{2}-\frac{1}{4} \sin (-2 \pi)\right)= \\
& =\frac{81}{8}\left(\frac{\pi}{2}-\frac{1}{4} \cdot 0\right)-\frac{81}{8}\left(-\frac{\pi}{2}-\frac{1}{4} \cdot 0\right)=\frac{81 \pi}{16}+\frac{81 \pi}{16}=\frac{81 \pi}{8}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-20$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

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- Last edited: 07:20, 2 July 2010.
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## Problem Kuznetsov Integrals 12-21

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