## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{1} x^{2} \sqrt{1-x^{2}} d x
$$
## Solution

$$
\int_{0}^{1} x^{2} \sqrt{1-x^{2}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=\sin t \Rightarrow d x=\cos t d t \\
& x=0 \Rightarrow t=\arcsin 0=0 \\
& x=1 \Rightarrow t=\arcsin 1=\frac{\pi}{2}
\end{aligned}
$$

We get:

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

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 $\% \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} \_12-2 »

Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals | Problems for Checking

- Last modified: 08:56, 9 July 2009.
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Kuznetsov's Problem Book Integrals 12-2 — PlusPi

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## Problem Kuznetsov Integrals 12-3

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