## Problem Statement

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

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

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

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

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-7$ »

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

- Last edited: 18:17, 28 April 2009.
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