## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{4}} \frac{2 \cos x + 3 \sin x}{(2 \sin x - 3 \cos x)^{3}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{2 \cos x+3 \sin x}{(2 \sin x-3 \cos x)^{3}} d x=\int_{0}^{\frac{\pi}{4}} \frac{d(2 \sin x-3 \cos x)}{(2 \sin x-3 \cos x)^{3}}=-\left.\frac{1}{2 \cdot(2 \sin x-3 \cos x)^{2}}\right|_{0} ^{\frac{\pi}{4}}= \\
& =-\frac{1}{2 \cdot\left(2 \sin \frac{\pi}{4}-3 \cos \frac{\pi}{4}\right)^{2}}+\frac{1}{2 \cdot(2 \sin 0-3 \cos 0)^{2}}= \\
& =-\frac{1}{2 \cdot\left(2 \cdot \frac{\sqrt{2}}{2}-3 \cdot \frac{\sqrt{2}}{2}\right)^{2}}+\frac{1}{2 \cdot(2 \cdot 0-3 \cdot 1)^{2}}= \\
& =-\frac{1}{2 \cdot\left(-\frac{\sqrt{2}}{2}\right)^{2}}+\frac{1}{2 \cdot(-3)^{2}}=-\frac{1}{2 \cdot \frac{2}{4}}+\frac{1}{2 \cdot(-3)^{2}}=-1+\frac{1}{18}=-\frac{17}{18}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-6 \%$

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

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## Problem Kuznetsov Integrals 4-7

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