## Condition of the problem

Calculate the definite integral:

$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{4 \operatorname{tg} x-5}{1-\sin 2 x+4 \cos ^{2} x} d x
$$
## Solution

$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{4 \operatorname{tg} x-5}{1-\sin 2 x+4 \cos ^{2} x} d x
$$

We will use the substitution:

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

From which:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}} \\
& \cos ^{2} x=\frac{1}{t^{2}+1} \\
& d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\operatorname{arctg} 3 \Rightarrow t=\operatorname{tg}(\operatorname{arctg} 3)=3
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{\pi / 4}^{\operatorname{arctg} 3} \frac{4 \operatorname{tg} x-5}{1-\sin 2 x+4 \cos ^{2} x} d x= \\
& =\int_{1}^{3} \frac{(4 t-5) \frac{d t}{1+t^{2}}}{1-\frac{2 t}{1+t^{2}}+\frac{4}{t^{2}+1}}=\int_{1}^{3} \frac{(4 t-5) d t}{1+t^{2}-2 t+4}= \\
& =\int_{1}^{3} \frac{4(t-1)-1}{(t-1)^{2}+4} d t=2 \int_{1}^{3} \frac{d\left((t-1)^{2}+4\right)}{(t-1)^{2}+4} d t-\int_{1}^{3} \frac{1}{(t-1)^{2}+4} d t
\end{aligned}
$$

$$
=\left.\left(2 \ln \left|(t-1)^{2}+4\right|-\frac{1}{2} \operatorname{arctg} \frac{t-1}{2}\right)\right|_{1} ^{3}=2 \ln 8-\frac{\pi}{8}-2 \ln 4=2 \ln 2-\frac{\pi}{8}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-4$ »

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

- Last edited: 09:29, July 7, 2010.
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Created by Geeteatoo

## Problem Kuznetsov Integrals 9-5

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