## Problem Statement

Calculate the definite integral:

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

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

Substitution:

$$
\begin{aligned}
& x=5 \sin t \Rightarrow d x=5 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{5}=0 \\
& x=\frac{5}{2} \Rightarrow t=\arcsin \frac{\left(\frac{5}{2}\right)}{5}=\arcsin \frac{1}{2}=\frac{\pi}{6}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 6} \frac{25 \sin ^{2} t \cdot 5 \cos t d t}{\sqrt{25-25 \sin ^{2} t}}=\int_{0}^{\pi / 6} \frac{25 \sin ^{2} t \cdot 5 \cos t}{5 \sqrt{1-\sin ^{2} t}} d t=25 \cdot \int_{0}^{\pi / 6} \frac{\sin ^{2} t \cdot \cos t}{\cos t} d t= \\
& =25 \cdot \int_{0}^{\pi / 6} \sin ^{2} t d t=25 \cdot \int_{0}^{\pi / 6} \frac{1-\cos 2 t}{2} d t=\frac{25}{2} \cdot \int_{0}^{\pi / 6}(1-\cos 2 t) d t= \\
& =\left.\frac{25}{2}\left(t-\frac{1}{2} \sin 2 t\right)\right|_{0} ^{\pi / 6}=\frac{25}{2}\left(\frac{\pi}{6}-\frac{1}{2} \sin \frac{\pi}{3}\right)-\frac{25}{2}\left(0-\frac{1}{2} \sin 0\right)= \\
& =\frac{25}{2}\left(\frac{\pi}{6}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}\right)-0=\frac{25 \pi}{12}-\frac{25 \sqrt{3}}{8}
\end{aligned}
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_12-14" Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

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Created by GeeTeatoo

## Problem Kuznetsov Integrals 12-15

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