## Solution

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{\sin x-\cos x}{(\cos x+\sin x)^{5}} d x=-\int_{0}^{\frac{\pi}{4}} \frac{d(\cos x+\sin x)}{(\cos x+\sin x)^{5}}=\left.\frac{1}{4} \cdot \frac{1}{(\cos x+\sin x)^{4}}\right|_{0} ^{\frac{\pi}{4}}= \\
& =\frac{1}{4} \cdot \frac{1}{\left(\cos \frac{\pi}{4}+\sin \frac{\pi}{4}\right)^{4}}-\frac{1}{4} \cdot \frac{1}{(\cos 0+\sin 0)^{4}}= \\
& =\frac{1}{4} \cdot \frac{1}{\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)^{4}}-\frac{1}{4} \cdot \frac{1}{(1+0)^{4}}=\frac{1}{4} \cdot \frac{1}{(\sqrt{2})^{4}}-\frac{1}{4}=\frac{1}{4} \cdot \frac{1}{4}-\frac{1}{4}= \\
& =\frac{1}{16}-\frac{4}{16}=-\frac{3}{16}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_4-27$ »

Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited: 10:09, March 23, 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 4-28

## Material from Pluspi
## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{4}} \frac{\sin x - \cos x}{(\cos x + \sin x)^{5}} d x
$$