## Problem Statement

Calculate the definite integral:

$$
\int_{6}^{10} \sqrt{\frac{4-x}{x-12}} d x
$$
## Solution

$$
\int_{6}^{10} \sqrt{\frac{4-x}{x-12}} d x=
$$

Substitution:

$$
\begin{aligned}
& t=\sqrt{\frac{4-x}{x-12}} \Rightarrow t^{2}=\frac{4-x}{x-12}=-\frac{x-12+8}{x-12}=-1-\frac{8}{x-12} \Rightarrow \\
& \Rightarrow 12-x=\frac{8}{t^{2}+1} \Rightarrow x=-\frac{8}{t^{2}+1}+12 \\
& d x=\frac{8}{\left(t^{2}+1\right)^{2}} \cdot 2 t \cdot d t=\frac{16 t}{\left(t^{2}+1\right)^{2}} \cdot d t \\
& x=6 \Rightarrow t=\sqrt{\frac{4-6}{6-12}}=\sqrt{\frac{-2}{-6}}=\frac{1}{\sqrt{3}} \\
& x=10 \Rightarrow t=\sqrt{\frac{4-10}{10-12}}=\sqrt{\frac{-6}{-2}}=\sqrt{3}
\end{aligned}
$$

We get:

$$
=\int_{\frac{1}{\sqrt{3}}}^{\sqrt{3}} t \cdot \frac{16 t}{\left(t^{2}+1\right)^{2}} \cdot d t=
$$

Substitution:

$$
\begin{aligned}
& t=\tan z, d t=\frac{d z}{\cos ^{2} z} \\
& t=\frac{1}{\sqrt{3}} \Rightarrow z=\arctan \frac{1}{\sqrt{3}}=\frac{\pi}{6} \\
& t=\sqrt{3} \Rightarrow z=\arctan \sqrt{3}=\frac{\pi}{3}
\end{aligned}
$$

We get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+11-11$ » Categories: Kuznetsov's Problem Book Integrals Problem 11 | Integrals

- Last edited: 17:29, 25 June 2009.
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## Problem Kuznetsov Integrals 11-12

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