## Problem Statement

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substituting:

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

Substitution:

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

We get:

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

$$
=-\frac{1}{u}-\left.\ln |u|\right|_{1} ^{2}=-\frac{1}{2}-\ln |2|+\frac{1}{1}+\ln |1|=\frac{1}{2}-\ln 2
$$

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_8-23» Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

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## Problem Kuznetsov Integrals 8-24

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