## Condition of the problem

Calculate the definite integral:

$$
\int_{\pi / 4}^{\operatorname{arctg} 3} \frac{1+\operatorname{ctg} x}{(\sin x+2 \cos x)^{2}} d x
$$
## Solution

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

We will use the substitution:

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

From which:

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

Substitute:

$$
\begin{aligned}
& =\int_{1}^{3} \frac{1+\frac{1}{t}}{1+\frac{4 t}{1+t^{2}}+\frac{3}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{1}^{3} \frac{\frac{t+1}{t}}{1+t^{2}+4 t+3} d t= \\
& =\int_{1}^{3} \frac{1+t}{t\left(t^{2}+4 t+4\right)} d t=\int_{1}^{3} \frac{1+t}{t(t+2)^{2}} d t=
\end{aligned}
$$

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

$$
\frac{1+t}{t(t+2)^{2}}=\frac{A}{t}+\frac{B_{1}}{t+2}+\frac{B_{2}}{(t+2)^{2}}=\frac{A(t+2)^{2}+B_{1} t(t+2)+B_{2} t}{t(t+2)^{2}}=
$$

$$
\begin{aligned}
& =\frac{A\left(t^{2}+4 t+4\right)+B_{1}\left(t^{2}+2 t\right)+B_{2} t}{t(t+2)^{2}}=\frac{\left(A+B_{1}\right) t^{2}+\left(4 A+2 B_{1}+B_{2}\right) t+4 A}{t(t+2)^{2}} \\
& \left\{\begin{array}{l}
A+B_{1}=0 \\
4 A+2 B_{1}+B_{2}=1 \\
4 A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=-\frac{1}{4} \\
2 B_{1}+B_{2}=0 \\
A=\frac{1}{4}
\end{array}\right. \\
& \left\{\begin{array}{l}
B_{1}=-\frac{1}{4} \\
B_{2}=\frac{1}{2} \\
A=\frac{1}{4}
\end{array}\right. \\
& \frac{1+t}{t(t+2)^{2}}=\frac{1}{4 t}-\frac{1}{4(t+2)}+\frac{1}{2(t+2)^{2}} \\
& =\int_{1}^{3}\left(\frac{1}{4 t}-\frac{1}{4(t+2)}+\frac{1}{2(t+2)^{2}}\right) d t=\left.\left(\frac{1}{4} \ln |t|-\frac{1}{4} \ln |t+2|-\frac{1}{2(t+2)}\right)\right|_{1} ^{3}= \\
& =\left(\frac{1}{4} \ln |3|-\frac{1}{4} \ln |3+2|-\frac{1}{2(3+2)}\right)-\left(\frac{1}{4} \ln |1|-\frac{1}{4} \ln |1+2|-\frac{1}{2(1+2)}\right)= \\
& =\frac{\ln 3}{4}-\frac{\ln 5}{4}-\frac{1}{10}+\frac{\ln 3}{4}+\frac{1}{6}=\frac{1}{4} \ln \frac{9}{5}+\frac{1}{15}
\end{aligned}
$$

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}+9-10 »

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

- Last edited: 20:45, 11 May 2009.
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## Problem Kuznetsov Integrals 9-11

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