## Problem Statement

Calculate the definite integral:

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

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

Substitution:

$$
\begin{aligned}
& u=1+t, d t=d u \\
& t=0 \Rightarrow u=1 \\
& t=1 \Rightarrow u=2
\end{aligned}
$$

We get:

$$
=\int_{1}^{2} \frac{(u-1) d u}{u^{2}}=\int_{1}^{2} \frac{d u}{u}-\int_{1}^{2} \frac{d u}{u^{2}}=\ln \left|u \right|_{1}^{2}+\frac{1}{u}\left|_{1}^{2} \right|=
$$

$$
=\ln |2|-\ln |1|+\frac{1}{2}-\frac{1}{1}=\ln 2-\frac{1}{2}
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_8-22 » Category: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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## Problem Kuznetsov Integrals 8-23

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