## Condition of the problem

Calculate the definite integral:

$$
\int_{0}^{\pi / 4} \frac{5 \operatorname{tg} x+2}{2 \sin 2 x+5} d x
$$
## Solution

Let's use the substitution:

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

From which:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{\pi / 4} \frac{5 \operatorname{tg} x+2}{2 \sin 2 x+5} d x=\int_{0}^{1} \frac{5 t+2}{\frac{4 t}{1+t^{2}}+5} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{1} \frac{5 t+2}{4 t+5+5 t^{2}} d t= \\
& =\frac{1}{2} \cdot \int_{0}^{1} \frac{10 t+4}{5 t^{2}+4 t+5} d t=\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(5 t^{2}+4 t+5\right)}{5 t^{2}+4 t+5}=\left.\frac{1}{2} \cdot \ln \left(5 t^{2}+4 t+5\right)\right|_{0} ^{1}= \\
& =\frac{1}{2} \cdot \ln \left(5 \cdot 1^{2}+4 \cdot 1+5\right)-\frac{1}{2} \cdot \ln \left(5 \cdot 0^{2}+4 \cdot 0+5\right)= \\
& =\frac{1}{2} \cdot \ln 14-\frac{1}{2} \cdot \ln 5=\frac{1}{2} \ln \frac{14}{5}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-20$ »

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

- Last modified: 12:02, 13 May 2009.
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Kuznetsov's Problem Book Integrals 9-20-PlusPi
Created by GeeTeatoo

Kuznetsov's Problem Book Integrals 9-20 — PlusPi
Created by GeeTeatoo

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## Kuznetsov's Problem Book Integrals $9-21$

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