## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{\pi}{4}} \operatorname{tg} x \cdot \ln (\cos x) d x
$$
## Solution

$$
\int_{0}^{\frac{\pi}{4}} \operatorname{tg} x \cdot \ln (\cos x) d x=
$$

Substitution:

$$
\begin{aligned}
& y=\ln \cos x \\
& d y=\frac{1}{\cos x} \cdot(-\sin x) \cdot d x=-\operatorname{tg} x \cdot d x \\
& x=0 \Rightarrow y=\ln \cos 0=\ln 1=0 \\
& x=\frac{\pi}{4} \Rightarrow y=\ln \cos \frac{\pi}{4}=\ln \frac{\sqrt{2}}{2}
\end{aligned}
$$

We get:

$$
=\int_{0}^{\ln \frac{\sqrt{2}}{2}}-y \cdot d y=-\left.\frac{y^{2}}{2}\right|_{0} ^{\ln \frac{\sqrt{2}}{2}}=-\frac{\left(\ln \frac{\sqrt{2}}{2}\right)^{2}}{2}+\frac{0^{2}}{2}=-\frac{1}{2} \cdot \ln ^{2} \frac{\sqrt{2}}{2}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-23$ »

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

- Last edited on this page: 19:54, 3 March 2009.
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## Problem Kuznetsov Integrals 4-24

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