## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2} \frac{x^{2} \cdot d x}{\sqrt{16-x^{2}}}
$$
## Solution

$$
\int_{0}^{2} \frac{x^{2} \cdot d x}{\sqrt{16-x^{2}}}=
$$

Substitution:

$$
\begin{aligned}
& x=4 \sin t \Rightarrow d x=4 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{4}=0 \\
& x=2 \Rightarrow t=\arcsin \frac{2}{4}=\frac{\pi}{6}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \frac{16 \sin ^{2} t \cdot 4 \cos t d t}{\sqrt{16-16 \sin ^{2} t}}=\int_{0}^{\pi / 6} \frac{16 \sin ^{2} t \cdot 4 \cos t}{4 \sqrt{1-\sin ^{2} t}} d t=16 \cdot \int_{0}^{\pi / 6} \frac{\sin ^{2} t \cdot \cos t}{\cos t} d t= \\
& =16 \cdot \int_{0}^{\pi / 6} \sin ^{2} t d t=16 \cdot \int_{0}^{\pi / 6} \frac{1-\cos 2 t}{2} d t=8 \cdot \int_{0}^{\pi / 6}(1-\cos 2 t) d t= \\
& =\left.8\left(t-\frac{1}{2} \sin 2 t\right)\right|_{0} ^{\pi / 6}=8\left(\frac{\pi}{6}-\frac{1}{2} \sin \frac{\pi}{3}\right)-8\left(0-\frac{1}{2} \sin 0\right)= \\
& =8\left(\frac{\pi}{6}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\right)-0=\frac{4 \pi}{3}-2 \sqrt{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+12-10$ » Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

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## Problem Kuznetsov Integrals 12-11

## Material from PlusPi

