## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\arcsin \sqrt{3 / 7}} \frac{\tan^{2} x \, dx}{3 \sin ^{2} x + 4 \cos ^{2} x - 7}
$$
## Solution

Let's use the substitution:

$$
t=\operatorname{tg} x
$$

From which we get:

$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\arcsin \sqrt{\frac{3}{7}} \Rightarrow t=\operatorname{tg}\left(\arcsin \sqrt{\frac{3}{7}}\right)=\sqrt{\frac{1}{1-\sin ^{2}\left(\arcsin \sqrt{\frac{3}{7}}\right)}-1}= \\
& =\sqrt{\frac{1}{1-\frac{3}{7}}-1}=\sqrt{\frac{7}{4}-1}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{\arcsin \sqrt{3 / 7}} \frac{\operatorname{tg}^{2} x d x}{3 \sin ^{2} x+4 \cos ^{2} x-7}=\int_{0}^{\sqrt{3} / 2} \frac{t^{2}}{\frac{3 t^{2}}{1+t^{2}}+\frac{4}{1+t^{2}}-7} \cdot \frac{d t}{1+t^{2}}= \\
= & \int_{0}^{\sqrt{3} / 2} \frac{t^{2}}{3 t^{2}+4-7-7 t^{2}} d t=\int_{0}^{\sqrt{3} / 2} \frac{t^{2}}{-3-4 t^{2}} d t=-\frac{1}{4} \cdot \int_{0}^{\sqrt{2}} \frac{t^{2}}{t^{2}+\frac{3}{4}} d t= \\
= & -\frac{1}{4} \cdot \int_{0}^{\sqrt{3} / 2} \frac{t^{2}+\frac{3}{4}-\frac{3}{4}}{t^{2}+\frac{3}{4}} d t=-\frac{1}{4} \cdot \int_{0}^{\sqrt{3} / 2} d t+\frac{3}{16} \cdot \int_{0}^{1} \frac{1}{t^{2}+\frac{3}{4}} d t=
\end{aligned}
$$

$$
\begin{aligned}
& =-\left.\frac{t}{4}\right|_{0} ^{\sqrt{3} / 2}+\left.\frac{3}{16} \cdot \frac{1}{\sqrt{\frac{3}{4}}} \cdot \operatorname{arctg} \frac{t}{\sqrt{\frac{3}{4}}}\right|_{0} ^{\sqrt{3} / 2}= \\
& =-\frac{\left(\frac{\sqrt{3}}{2}\right)}{4}+\frac{0}{4}+\frac{\sqrt{3}}{8} \cdot \operatorname{arctg} \frac{2 \cdot \frac{\sqrt{3}}{2}}{\sqrt{3}}-\frac{\sqrt{3}}{8} \cdot \operatorname{arctg} \frac{2 \cdot 0}{\sqrt{3}}= \\
& =-\frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{8} \cdot \operatorname{arctg} 1-0=-\frac{\sqrt{3}}{8}+\frac{\sqrt{3}}{8} \cdot \frac{\pi}{4}=-\frac{\sqrt{3}}{8}+\frac{\sqrt{3} \pi}{32}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-16 \%$ Categories: Kuznetsov Problem Book Integrals Problem 9 | Integrals

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## Problem Kuznetsov Integrals 9-17

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