## Problem Statement

Calculate the definite integral:

$$
\int_{-\frac{4}{\sqrt{17}}}^{\pi / 4} \frac{2 \operatorname{ctg} x+1}{(2 \sin x+\cos x)^{2}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{\arccos \frac{4}{\sqrt{17}}}^{\frac{\pi}{4}} \frac{2 \operatorname{ctg} x+1}{(2 \sin x+\cos x)^{2}} d x=\int_{\arccos \frac{4}{\sqrt{17}}}^{\frac{\pi}{4}} \frac{2 \operatorname{tg} x+1}{(2 \operatorname{tg} x+1)^{2} \cdot \cos ^{2} x} d x=\left|d(\operatorname{tg} x)=\frac{d x}{\cos ^{2} x}\right|= \\
& =\int^{\frac{\pi}{4}} \frac{2 \operatorname{ctg} x+1}{(2 \operatorname{tg} x+1)^{2}} d(\operatorname{tg} x)
\end{aligned}
$$

Substitution:

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

$$
\begin{aligned}
& \text { When } x=\arccos \frac{4}{\sqrt{17}}, \quad t=\frac{1}{4} \\
& \text { When } x=\frac{\pi}{4}, \quad t=1
\end{aligned}
$$

We have:

$$
\begin{aligned}
& \int_{\arccos \frac{4}{\sqrt{17}}}^{\frac{\pi}{4}} \frac{2 \operatorname{ctg} x+1}{(2 \operatorname{tg} x+1)^{2}} d(\operatorname{tg} x)=\int_{\frac{1}{4}}^{1} \frac{\frac{2}{t}+1}{(2 t+1)^{2}} d t=\int_{\frac{1}{4}}^{1} \frac{t+2}{t(2 t+1)^{2}} d t \\
& \frac{t+2}{t(2 t+1)^{2}}=\frac{A}{t}+\frac{B}{2 t+1}+\frac{C}{(2 t+1)^{2}}=\frac{A(2 t+1)^{2}+B t(2 t+1)+C t}{t(2 t+1)^{2}} \\
& t+2=A(2 t+1)^{2}+B t(2 t+1)+C t \\
& t=0 \Rightarrow A=2 \\
& t=-\frac{1}{2} \Rightarrow-\frac{1}{2} C=-\frac{1}{2}+2 ; C=-3 \\
& t=-1 \Rightarrow A+B-C=1 ; 2+B+3=1 ; \quad B=-4 \\
& \int_{\frac{1}{4}}^{1} \frac{t+2}{t(2 t+1)^{2}} d t=\int_{\frac{1}{4}}^{1}\left(\frac{2}{4}-\frac{4}{2 t+1}-\frac{3}{(2 t+1)^{2}} d t\right)=\left.\left(2 \ln |t|-2 \ln |2 t+1|+\frac{3}{2} \cdot \frac{1}{2 t+1}\right)\right|_{\frac{1}{4}} ^{1} \\
& =\left.\left(2 \ln \left|\frac{t}{2 t+1}\right|+\frac{3}{2} \cdot \frac{1}{2 t+1}\right)\right|_{\frac{1}{4}} ^{1}=2 \ln \frac{1}{3}+\frac{1}{2}-2 \ln \frac{1}{6}-1=2 \ln 2-\frac{1}{2}
\end{aligned}
$$

$\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-2 »$

Categories: Kuznetsov Integral Problems Problem 9| Integrals

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## Problem Kuznetsov Integrals 9-3

## Material from PlusPi

This problem may have a different condition (possibly due to different editions or errors). For more details, see 9-3(2)