## Problem Statement

Calculate the definite integral:

$$
\int_{\arcsin (1 / \sqrt{37})}^{\pi / 4} \frac{6 \operatorname{tg} x d x}{3 \sin 2 x+5 \cos ^{2} x}
$$
## Solution

$\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorname{tg} x d x}{3 \sin 2 x+5 \cos ^{2} x}=\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorname{tg} x d x}{6 \sin x \cdot \cos x+5 \cos ^{2} x}=$

$=\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorname{tg} x d x}{(6 \operatorname{tg} x+5) \cos ^{2} x}=\int_{\arcsin \frac{1}{\sqrt{37}}}^{\frac{\pi}{4}} \frac{6 \operatorname{tg} x d(\operatorname{tg} x)}{6 \operatorname{tg} x+5}=\left|\begin{array}{|l|l|l}t=\operatorname{tg} x & \arcsin \frac{1}{\sqrt{37}} & \frac{\pi}{4} \\ \hline t & \frac{1}{6} & 1\end{array}\right|=$

$=\int_{\frac{1}{6}}^{1} \frac{6 \mathrm{tdt}}{6 \mathrm{t}+5}=\int_{\frac{1}{6}}^{1} \frac{6 \mathrm{t}+5-5}{6 \mathrm{t}+5} \mathrm{dt}=\int_{\frac{1}{6}}^{1}\left(1-\frac{5}{6 \mathrm{t}+5}\right) \mathrm{dt}=$

$=\left.\left(t-\frac{5}{6} \ln |6 t+5|\right)\right|_{\frac{1}{6}} ^{1}=1-\frac{5}{6} \ln 11-\frac{1}{6}+\frac{5}{6} \ln 6=\frac{5}{6}(1+\ln 6-\ln 11)=$

$=\frac{5}{6}\left(1+\ln \frac{6}{11}\right)=\frac{5}{6} \ln \frac{6 \cdot \mathrm{e}}{11}$.

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Categories: In the form of a scan | Kuznetsov's Problem Book Integrals Problem $9 \mid$ Integrals

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## Problem Kuznetsov Integrals $9-8$

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