## Problem Statement

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

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

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

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

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

Subtract the first equation from the third:

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

Add the third equation to the second:

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

Then:

$$
\begin{aligned}
& =\int_{0}^{1}\left(\frac{1}{1+t}-\frac{t}{1+t^{2}}+\frac{1}{1+t^{2}}\right) d t=\int_{0}^{1} \frac{d t}{1+t}-\frac{1}{2} \cdot \int_{0}^{1} \frac{d\left(t^{2}+1\right)}{1+t^{2}}+\int_{0}^{1} \frac{1}{1+t^{2}} d t= \\
& =\left.\ln |1+t|\right|_{0} ^{1}-\frac{1}{2} \cdot \ln \left|1+t^{2} \|_{0}^{1}+\operatorname{arctg} t\right|_{0}^{1}= \\
& =\ln |1+1|-\ln |1+0|-\frac{1}{2} \cdot \ln \left|1+1^{2}\right|+\frac{1}{2} \cdot \ln \left|1+0^{2}\right|+\operatorname{arctg} 1-\operatorname{arctg} 0= \\
& =\ln 2-0-\frac{1}{2} \cdot \ln 2+0+\frac{\pi}{4}-0=\frac{1}{2} \cdot \ln 2+\frac{\pi}{4}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_8-12$ »

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

- Last edited: May 1, 2009, at 09:52.
- Content is available under CC-BY-SA 3.0.