## Condition of the problem

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{2 \operatorname{arctg} \frac{1}{2}}^{\frac{\pi}{2}} \frac{\cos x d x}{(1-\cos x)^{3}}=\int_{\frac{1}{2}}^{1} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{\left(1-\frac{1-t^{2}}{1+t^{2}}\right)^{3}}=\int_{\frac{1}{2}}^{1} \frac{2\left(1-t^{2}\right) d t}{\left(1+t^{2}\right)^{2} \cdot\left(\frac{1+t^{2}-1+t^{2}}{1+t^{2}}\right)^{3}}= \\
= & \int_{\frac{1}{2}}^{1} \frac{2\left(1-t^{2}\right) d t}{\left(\frac{8 t^{6}}{1+t^{2}}\right)}=\int_{\frac{1}{2}}^{1^{2}} \frac{\left(1+t^{2}\right)\left(1-t^{2}\right) d t}{4 t^{6}}=\frac{1}{4} \cdot \int_{\frac{1}{2}}^{1} \frac{\left(1-t^{4}\right) d t}{t^{6}}= \\
= & \frac{1}{4} \cdot \int_{\frac{1}{2}}^{1}\left(\frac{1}{t^{6}}-\frac{1}{t^{2}}\right) d t=\left.\frac{1}{4} \cdot\left(-\frac{1}{5 t^{5}}+\frac{1}{t}\right)\right|_{\frac{1}{2}} ^{1}= \\
= & \frac{1}{4} \cdot\left(\left(-\frac{1}{5 \cdot 1^{5}}+\frac{1}{1}\right)-\left(-\frac{1}{5 \cdot\left(\frac{1}{2}\right)^{5}}+\frac{1}{\left(\frac{1}{2}\right)}\right)\right)=
\end{aligned}
$$

$$
=\frac{1}{4} \cdot\left(-\frac{1}{5}+1+\frac{32}{5}-2\right)=\frac{1}{4} \cdot\left(\frac{31}{5}-1\right)=\frac{1}{4} \cdot \frac{26}{5}=\frac{13}{10}=1.3
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-4$ » Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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## Problem Kuznetsov Integrals $8-5$

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