## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{4} \sqrt{16-x^{2}} d x
$$
## Solution

$$
\int_{0}^{4} \sqrt{16-x^{2}} d x=
$$

Substitution:

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

We get:

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

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_12-16" Categories: Kuznetsov's Problem Book Integrals Problem 12 | Integrals

- Last edited on this page: 16:06, 5 July 2009.
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## Problem Kuznetsov Integrals 12-17

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