## Condition of the problem

Calculate the definite integral:

$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{2}} \frac{1+\sin x}{(1-\sin x)^{2}} d x
$$
## Solution

We will use the universal substitution:

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

From which:

$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=2 \operatorname{arctg} \frac{1}{2} \Rightarrow t=\operatorname{tg} \frac{2 \operatorname{arctg} \frac{1}{2}}{2}=\operatorname{tg}\left(2 \operatorname{arctg} \frac{1}{2}\right)=\frac{1}{2}
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{2 \operatorname{arctg} \frac{1}{2}} \frac{1+\sin x}{(1-\sin x)^{2}} d x=\int_{0}^{\frac{1}{2}} \frac{\left(1+\frac{2 t}{1+t^{2}}\right) \cdot \frac{2 d t}{1+t^{2}}}{\left(1-\frac{2 t}{1+t^{2}}\right)^{2}}=\int_{0}^{\frac{1}{2}} \frac{\frac{1+t^{2}+2 t}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{\left(\frac{1+t^{2}-2 t}{1+t^{2}}\right)^{2}}= \\
& =2 \cdot \int_{0}^{\frac{1}{2}} \frac{(1+t)^{2} d t}{(1-t)^{4}}=
\end{aligned}
$$

Substitution:

$$
\begin{aligned}
& u=1-t, d t=-d u \\
& t=0 \Rightarrow u=1 \\
& t=\frac{1}{2} \Rightarrow u=\frac{1}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =-2 \cdot \int_{1}^{\frac{1}{2}} \frac{(2-u)^{2}}{u^{4}} d u=-2 \cdot \int_{1}^{\frac{1}{2}} \frac{4-4 u+u^{2}}{u^{4}} d u= \\
& =2 \cdot \int_{1}^{\frac{1}{2}}\left(\frac{4}{u^{4}}-\frac{4}{u^{3}}+\frac{1}{u^{2}}\right) d u=-\left.2 \cdot\left(-\frac{4}{3 u^{3}}+\frac{2}{u^{2}}-\frac{1}{u}\right)\right|_{1} ^{\frac{1}{2}}= \\
& -2\left(-\frac{4}{3\left(\frac{1}{2}\right)^{3}}+\frac{2}{\left(\frac{1}{2}\right)^{2}}-\frac{1}{\left(\frac{1}{2}\right)}\right)+2\left(-\frac{4}{3 \cdot 1^{3}}+\frac{2}{1^{2}}-\frac{1}{1}\right)= \\
& =-2\left(-\frac{32}{3}+8-2\right)+2\left(-\frac{4}{3}+2-1\right)=\frac{64}{3}-16+4-\frac{8}{3}+4-2= \\
& =\frac{56}{3}-10=\frac{26}{3}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-14$ »

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

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## Problem Kuznetsov Integrals $8-15$

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