## Condition of the problem

Calculate the definite integral:

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

$$
\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{x \cdot \cos x+\sin x}{(x \cdot \sin x)^{2}} d x=
$$

Substitution:

$$
\begin{aligned}
& y=x \cdot \sin x \\
& d y=(\sin x+x \cdot \cos x) d x \\
& x=\frac{\pi}{4} \Rightarrow y=\frac{\pi}{4} \cdot \sin \frac{\pi}{4}=\frac{\sqrt{2} \pi}{8} \\
& x=\frac{\pi}{2} \Rightarrow y=\frac{\pi}{2} \cdot \sin \frac{\pi}{2}=\frac{\pi}{2}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{\frac{\sqrt{2} \pi}{8}}^{\frac{\pi}{2}} \frac{d y}{y^{2}}=-\left.\frac{1}{y}\right|_{\frac{\sqrt{2} \pi}{8}} ^{\frac{\pi}{2}}=-\frac{1}{\left(\frac{\pi}{2}\right)}+\frac{1}{\left(\frac{\sqrt{2} \pi}{8}\right)}= \\
& =-\frac{2}{\pi}+\frac{8}{\sqrt{2} \pi}=-\frac{2}{\pi}+\frac{4 \sqrt{2}}{\pi}=\frac{4 \sqrt{2}-2}{\pi}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-28$ »

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

- Last edited: 22:15, March 3, 2009.
- Content is available under CC-BY-SA 3.0.