## Condition of the problem

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

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

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

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

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

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80 $\% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-8 \%$

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

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## Problem Kuznetsov Integrals $8-9$

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