## Problem Statement

Calculate the definite integral:

$$
\int_{\arcsin (2 / \sqrt{5})}^{\arcsin (3 / \sqrt{10})} \frac{2 \tan x + 5}{(5 - \tan x) \sin 2x} \, dx
$$
## Solution

We will use the substitution:

$$
t=\operatorname{tg} x
$$

From which:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\arcsin \frac{2}{\sqrt{5}} \Rightarrow t=\operatorname{tg}\left(\arcsin \frac{2}{\sqrt{5}}\right)=\sqrt{\frac{1}{1-\sin ^{2}\left(\arcsin \frac{2}{\sqrt{5}}\right)}-1}= \\
& =\sqrt{\frac{1}{1-\frac{4}{5}}-1}=\sqrt{5-1}=\sqrt{4}=2 \\
& x=\arcsin \frac{3}{\sqrt{10}} \Rightarrow t=\operatorname{tg}\left(\arcsin \frac{3}{\sqrt{10}}\right)=\sqrt{\frac{1}{1-\sin ^{2}\left(\arcsin \frac{3}{\sqrt{10}}\right)}-1}= \\
& =\sqrt{\frac{1}{1-\frac{9}{10}}-1}=\sqrt{10-1}=\sqrt{9}=3
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{\arcsin (2 / \sqrt{5})}^{\arcsin (3 / \sqrt{10})} \frac{2 \operatorname{tg} x+5}{(5-\operatorname{tg} x) \sin 2 x} d x=\int_{2}^{3} \frac{2 t+5}{(5-t) \cdot \frac{2 t}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}= \\
& =-\frac{1}{2} \cdot \int_{2}^{3} \frac{2 t+5}{t(t-5)} d t=
\end{aligned}
$$

Decompose the proper rational fraction into elementary fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{2 t+5}{t(t-5)}=\frac{A}{t}+\frac{B}{t-5}=\frac{A(t-5)+B t}{t(t-5)}=\frac{(A+B) t-5 A}{t(t-5)} \\
& \left\{\begin{array}{l}
A+B=2 \\
-5 A=5
\end{array}\right. \\
& \left\{\begin{array}{l}
B=3 \\
A=-1
\end{array}\right. \\
& \frac{2 t+5}{t(t-5)}=-\frac{1}{t}+\frac{3}{t-5}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =-\frac{1}{2} \cdot \int_{2}^{3}\left(-\frac{1}{t}+\frac{3}{t-5}\right) d t=\frac{1}{2} \cdot \int_{2}^{3} \frac{1}{t} d t-\frac{3}{2} \cdot \int_{2}^{3} \frac{1}{t-5} d t= \\
& =\left.\frac{1}{2} \cdot \ln |t|\right|_{2} ^{3}-\left.\frac{3}{2} \cdot \ln |t-5|\right|_{2} ^{3}= \\
& =\frac{1}{2} \cdot \ln |3|-\frac{1}{2} \cdot \ln |2|-\frac{3}{2} \cdot \ln |3-5|+\frac{3}{2} \cdot \ln |2-5|= \\
& =\frac{1}{2} \cdot \ln \frac{3}{2}+\frac{3}{2} \cdot \ln \frac{3}{2}=2 \ln \frac{3}{2}
\end{aligned}
$$

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}+9-18 »

Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

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Kuznetsov's Problem Book Integrals 9-19

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