## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{16} \sqrt{256-x^{2}} d x
$$
## Solution

$$
\int_{0}^{16} \sqrt{256-x^{2}} d x=
$$

Substitution:

$$
\begin{aligned}
& x=16 \sin t \Rightarrow d x=16 \cos t d t \\
& x=0 \Rightarrow t=\arcsin \frac{0}{16}=0 \\
& x=16 \Rightarrow t=\arcsin \frac{16}{16}=\frac{\pi}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{0}^{\pi / 2} \sqrt{256-256 \sin ^{2} t} \cdot 16 \cos t d t=\int_{0}^{\pi / 2} 16 \cos t \cdot 16 \cos t d t= \\
& =256 \cdot \int_{0}^{\pi / 2} \cos ^{2} t d t=256 \cdot \int_{0}^{\pi / 2} \frac{1+\cos 2 t}{2} d t=128 \cdot \int_{0}^{\pi / 2}(1+\cos 2 t) d t= \\
& =\left.128\left(t+\frac{1}{2} \sin 2 t\right)\right|_{0} ^{\pi / 2}=128\left(\frac{\pi}{2}+\frac{1}{2} \sin \pi\right)-128\left(0+\frac{1}{2} \sin 0\right)= \\
& =128\left(\frac{\pi}{2}+\frac{1}{2} \cdot 0\right)-128\left(0+\frac{1}{2} \cdot 0\right)=64 \pi
\end{aligned}
$$

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

- Last modified: 08:56, 9 July 2009.
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Problem Kuznetsov Integrals 12-1 — PlusPi

Created by GeeTeatoo

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## Problem Kuznetsov Integrals 12-2

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