## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{2}} \frac{\cos x \, dx}{1+\sin x+\cos x}
$$
## 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:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D1} \% 8 \mathrm{~B}+8-15 »

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

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## Problem Kuznetsov Integrals 8-16

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