## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{5} x^{2} \cdot \sqrt{25-x^{2}} d x
$$
## Solution

$$
\int_{0}^{5} x^{2} \cdot \sqrt{25-x^{2}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=5 \sin t \Rightarrow d x=5 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{5}=0 \\
& x=5 \Rightarrow t=\arcsin \frac{5}{5}=\frac{\pi}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 2} 25 \sin ^{2} t \cdot \sqrt{25-25 \sin ^{2} t} \cdot 5 \cos t d t=\int_{0}^{\pi / 2} 25 \sin ^{2} t \cdot 5 \cos t \cdot 5 \cos t d t= \\
& =5^{4} \cdot \int_{0}^{\pi / 2} \sin ^{2} t \cdot \cos ^{2} t d t=\frac{5^{4}}{4} \cdot \int_{0}^{\pi / 2} \sin ^{2} 2 t d t=\frac{5^{4}}{4} \cdot \int_{0}^{\pi / 2} \frac{1-\cos 4 t}{2} d t= \\
& =\frac{5^{4}}{8} \cdot \int_{0}^{\pi / 2}(1-\cos 4 t) d t=\left.\frac{5^{4}}{8}\left(t-\frac{1}{4} \sin 4 t\right)\right|_{0} ^{\pi / 2}= \\
& =\frac{5^{4}}{8}\left(\frac{\pi}{2}-\frac{1}{4} \sin 2 \pi\right)-\frac{5^{4}}{8}\left(0-\frac{1}{4} \sin 0\right)=\frac{5^{4}}{8}\left(\frac{\pi}{2}-\frac{1}{4} \cdot 0\right)-\frac{5^{4}}{8}\left(0-\frac{1}{4} \cdot 0\right)=\frac{625 \pi}{16}
\end{aligned}
$$

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 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-15$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

- Last edited: 15:29, 5 July 2009.
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## Problem Kuznetsov Integrals 12-16

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