## Condition of the problem

Calculate the definite integral:

$$
\int_{0}^{\arccos \frac{1}{\sqrt{17}}} \frac{3+2 \operatorname{tg} x}{2 \sin ^{2} x+3 \cos ^{2} x-1} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\arccos \frac{1}{\sqrt{17}}} \frac{3+2 \tan x}{2 \sin ^{2} x+3 \cos ^{2} x-1} d x=\int_{0}^{\arccos } \frac{1}{\sqrt{17}} \frac{3+2 \tan x}{2+\cos ^{2} x-1} d x= \\
= & \int_{0}^{\frac{\arccos }{\sqrt{17}}} \frac{3+2 \tan x}{1+\cos ^{2} x} d x=
\end{aligned}
$$

We will use the substitution:

$$
t=\tan x
$$

From which:

$$
\begin{aligned}
& \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\tan 0=0 \\
& x=\arccos \frac{1}{\sqrt{17}} \Rightarrow t=\tan\left(\arccos \frac{1}{\sqrt{17}}\right)=\sqrt{\frac{1}{\cos ^{2}\left(\arccos \frac{1}{\sqrt{17}}\right)}-1}= \\
& =\sqrt{\frac{1}{\left(\frac{1}{17}\right)}-1}=\sqrt{17-1}=\sqrt{16}=4
\end{aligned}
$$

Substitute:

$$
=\int_{0}^{4} \frac{3+2 t}{1+\frac{1}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{4} \frac{3+2 t}{1+t^{2}+1} d t=\int_{0}^{4} \frac{3+2 t}{2+t^{2}} d t=
$$

$$
\begin{aligned}
& =3 \cdot \int_{0}^{4} \frac{1}{2+t^{2}} d t+\int_{0}^{4} \frac{d\left(t^{2}+2\right)}{2+t^{2}} d t=\left.3 \cdot \frac{1}{\sqrt{2}} \arctan \frac{t}{\sqrt{2}}\right|_{0} ^{4}+\left.\ln \left(t^{2}+2\right)\right|_{0} ^{4}= \\
& =\frac{3}{\sqrt{2}} \arctan \frac{4}{\sqrt{2}}-\frac{3}{\sqrt{2}} \arctan \frac{0}{\sqrt{2}}+\ln \left(4^{2}+2\right)-\ln \left(0^{2}+2\right)= \\
& =\frac{3}{\sqrt{2}} \arctan(2 \sqrt{2})+\ln (16+2)-\ln 2= \\
& =\frac{3}{\sqrt{2}} \arctan(2 \sqrt{2})+\ln \frac{18}{2}=\frac{3}{\sqrt{2}} \arctan(2 \sqrt{2})+\ln 9
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_9-3(2) »$ Categories: Kuznetsov's Problem Book Integrals Problem 9| Integrals

- Last edited: 10:46, 12 June 2009.
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## Problem Kuznetsov Integrals 9-4

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