## Problem Statement

Calculate the definite integral:

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

We will use the universal substitution:

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

From which:

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

Substitute:

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

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}+8-18 »

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

- Last modified: 18:58, 1 May 2009.
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## Problem Kuznetsov Integrals 8-19

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