## Problem Statement

Calculate the definite integral:

$$
\int_{2 \operatorname{arctan} 2}^{2 \operatorname{arctan} 3} \frac{d x}{\cos x(1-\cos x)}
$$
## Solution

We will use the universal substitution:

$$
t=\operatorname{tg} \frac{x}{2}
$$

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{2 \operatorname{arctg} 2}^{2 \operatorname{arctg} 3} \frac{d x}{\cos x(1-\cos x)}=\int_{2}^{3} \frac{\frac{2 d t}{1+t^{2}}}{\frac{1-t^{2}}{1+t^{2}} \cdot\left(1-\frac{1-t^{2}}{1+t^{2}}\right)}= \\
& =\int_{2}^{3} \frac{2 d t}{\left(1-t^{2}\right) \cdot \frac{1+t^{2}-1+t^{2}}{1+t^{2}}}=-\int_{2}^{3} \frac{2\left(1+t^{2}\right) d t}{\left(t^{2}-1\right) \cdot 2 t^{2}}=-\int_{2}^{3} \frac{1+t^{2}}{(t-1)(t+1) t^{2}} d t=
\end{aligned}
$$

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

$$
\begin{aligned}
& \frac{1+t^{2}}{(t-1)(t+1) t^{2}}=\frac{A_{1}}{t}+\frac{A_{2}}{t^{2}}+\frac{B}{t-1}+\frac{C}{t+1}= \\
& =\frac{A_{1}(t-1)(t+1) t+A_{2}(t-1)(t+1)+B(t+1) t^{2}+C(t-1) t^{2}}{(t-1)(t+1) t^{2}}= \\
& =\frac{A_{1}\left(t^{3}-t\right)+A_{2}\left(t^{2}-1\right)+B\left(t^{3}+t^{2}\right)+C\left(t^{3}-t^{2}\right)}{(t-1)(t+1) t^{2}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\left(A_{1}+B+C\right) t^{3}+\left(A_{2}+B-C\right) t^{2}-A_{1} t-A_{2}}{(t-1)(t+1) t^{2}}= \\
& \left\{\begin{array}{l}
A_{1}+B+C=0 \\
A_{2}+B-C=1 \\
-A_{1}=0 \\
-A_{2}=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B+C=0 \\
B-C=2 \\
A_{1}=0 \\
A_{2}=-1
\end{array}\right.
\end{aligned}
$$

Add the first equation to the second:

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=0 \\
2 B=2 \\
A_{1}=0 \\
A_{2}=-1
\end{array}\right. \\
& \left\{\begin{array}{l}
C=-1 \\
B=1 \\
A_{1}=0 \\
A_{2}=-1
\end{array}\right. \\
& \quad \frac{1+t^{2}}{(t-1)(t+1) t^{2}}=-\frac{1}{t^{2}}+\frac{1}{t-1}-\frac{1}{t+1} \\
& =-\int_{2}^{3}\left(-\frac{1}{t^{2}}+\frac{1}{t-1}-\frac{1}{t+1}\right) d t=\int_{2}^{3}\left(\frac{1}{t^{2}}-\frac{1}{t-1}+\frac{1}{t+1}\right) d t= \\
& =\left.\left(-\frac{1}{t}-\ln |t-1|+\ln |1+t|\right)\right|_{2} ^{3}= \\
& =\left(-\frac{1}{3}-\ln |3-1|+\ln |1+3|\right)-\left(-\frac{1}{2}-\ln |2-1|+\ln |1+2|\right)= \\
& =-\frac{1}{3}-\ln 2+\ln 4+\frac{1}{2}+\ln 1-\ln 3=\frac{1}{6}-\ln 2+2 \ln 2+0-\ln 3= \\
& =\frac{1}{6}+\ln 2-\ln 3
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-6$ »

Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

- Last edited: 16:55, 28 April 2009.
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## Problem Kuznetsov Integrals 8-7

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