## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x d x}{(1+\cos x+\sin x)^{2}}
$$
## Solution

We will use the universal substitution:

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

From which:

$$
\begin{aligned}
& \sin x=\frac{2 t}{1+t^{2}}, \cos x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{2 d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} \frac{0}{2}=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{2} \Rightarrow t=\operatorname{tg} \frac{\left(\frac{\pi}{2}\right)}{2}=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x d x}{(1+\cos x+\sin x)^{2}}=\int_{0}^{1} \frac{\left(\frac{2 t}{1+t^{2}}\right)^{2} \cdot \frac{2 d t}{1+t^{2}}}{\left(1+\frac{1-t^{2}}{1+t^{2}}+\frac{2 t}{1+t^{2}}\right)^{2}}=\int_{0}^{1} \frac{\left(\frac{2 t}{1+t^{2}}\right)^{2} \cdot \frac{2 d t}{1+t^{2}}}{\left(\frac{1+t^{2}+1-t^{2}+2 t}{1+t^{2}}\right)^{2}}= \\
& =\int_{0}^{1} \frac{8 t^{2} \cdot d t}{\left(1+t^{2}\right)(2+2 t)^{2}}=\int_{0}^{1} \frac{8 t^{2} \cdot d t}{4\left(1+t^{2}\right)(1+t)^{2}}=\int_{0}^{1} \frac{2 t^{2} \cdot d t}{\left(1+t^{2}\right)(1+t)^{2}}=
\end{aligned}
$$

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

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

$$
\left\{\begin{array}{l}
A+C_{1}=0 \\
2 A+B+C_{1}+C_{2}=2 \\
A+2 B+C_{1}=0 \\
B+C_{1}+C_{2}=0
\end{array}\right.
$$

Subtract the fourth equation from the second:

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

Then

$$
\begin{aligned}
& =\int_{0}^{1}\left(\frac{t}{1+t^{2}}-\frac{1}{1+t}+\frac{1}{(1+t)^{2}}\right) d t=\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(t^{2}+1\right)}{1+t^{2}}-\int_{0}^{1} \frac{d(t+1)}{1+t}+\int_{0}^{1} \frac{d(t+1)}{(1+t)^{2}}= \\
& =\left.\frac{1}{2} \cdot \ln \left(t^{2}+1\right)\right|_{0} ^{1}-\ln |t+1|_{0}^{1}-\left.\frac{1}{1+t}\right|_{0} ^{1}= \\
& =\frac{1}{2} \cdot \ln \left(1^{2}+1\right)-\frac{1}{2} \cdot \ln \left(0^{2}+1\right)-\ln |1+1|+\ln |0+1|-\frac{1}{1+1}+\frac{1}{1+0}= \\
& =\frac{1}{2} \cdot \ln 2-0-\ln 2+0-\frac{1}{2}+1=\frac{1}{2}-\frac{1}{2} \cdot \ln 2
\end{aligned}
$$

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\%D1\%82 $\% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \quad 8-25$ » Categories: Kuznetsov's Problem Book Integrals Problem $8 \mid$ Integrals

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## Problem Kuznetsov Integrals 8-26

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