## Problem Statement

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

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

Substitution:

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

We get:

$$
=\int_{1}^{2} \frac{4(u-1) \cdot d u}{u^{4}}=4 \cdot \int_{1}^{2}\left(\frac{1}{u^{3}}-\frac{1}{u^{4}}\right) d u=\left.4 \cdot\left(-\frac{1}{2 u^{2}}+\frac{1}{3 u^{3}}\right)\right|_{1} ^{2}=
$$

$$
\begin{aligned}
& =4 \cdot\left(-\frac{1}{2 \cdot 2^{2}}+\frac{1}{3 \cdot 2^{3}}\right)-4 \cdot\left(-\frac{1}{2 \cdot 1^{2}}+\frac{1}{3 \cdot 1^{3}}\right)= \\
& =-\frac{1}{2}+\frac{1}{6}+2-\frac{4}{3}=\frac{1}{3}
\end{aligned}
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_8-21 »

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

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- Last modified: 13:20, 2 May 2009.
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