## Problem Statement

Calculate the definite integral:

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

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+8-28$ »

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

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## Problem Kuznetsov Integrals $8-29$

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