## Condition of the problem

Calculate the definite integral:

$$
\int_{0}^{\pi / 4} \frac{7+3 \tan x}{(\sin x+2 \cos x)^{2}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\pi / 4} \frac{7+3 \tan x}{(\sin x+2 \cos x)^{2}} d x=\int_{0}^{\pi / 4} \frac{7+3 \tan x}{\sin ^{2} x+4 \sin x \cos x+4 \cos ^{2} x} d x= \\
& =\int_{0}^{\pi / 4} \frac{7+3 \tan x}{1+2 \sin 2 x+3 \cos ^{2} x} d x=
\end{aligned}
$$

We will use the substitution:

$$
t=\tan x
$$

From which:

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

Substitute:

$$
\begin{aligned}
& =\int_{0}^{1} \frac{7+3 t}{1+\frac{4 t}{1+t^{2}}+\frac{3}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{1} \frac{7+3 t}{1+t^{2}+4 t+3} d t= \\
& =\int_{0}^{1} \frac{7+3 t}{t^{2}+4 t+4} d t=\int_{0}^{1} \frac{3(t+2)+1}{(t+2)^{2}} d t=\int_{0}^{1} \frac{3}{t+2} d t+\int_{0}^{1} \frac{1}{(t+2)^{2}} d t= \\
& =3 \ln |t+2|_{0}^{1}-\left.\frac{1}{t+2}\right|_{0} ^{1}=3 \ln |1+2|-3 \ln |0+2|-\frac{1}{1+2}+\frac{1}{0+2}= \\
& =3 \ln 3-3 \ln 2-\frac{1}{3}+\frac{1}{2}=3 \ln \frac{3}{2}+\frac{1}{6}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-17$ »

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

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## Problem Kuznetsov Integrals 9-18

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