## Problem Statement

Calculate the definite integral:

$$
\int_{\pi / 4}^{\arcsin \sqrt{2 / 3}} \frac{8 \operatorname{tg} x d x}{3 \cos ^{2} x+8 \sin 2 x-7}
$$
## Solution

We will use the substitution:

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

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{\pi / 4}^{\arcsin \sqrt{2 / 3}} \frac{8 \operatorname{tg} x d x}{3 \cos ^{2} x+8 \sin 2 x-7}=\int_{1}^{\sqrt{2}} \frac{8 t}{\frac{3}{1+t^{2}}+\frac{16 t}{1+t^{2}}-7} \cdot \frac{d t}{1+t^{2}}= \\
= & \int_{1}^{\sqrt{2}} \frac{8 t}{3+16 t-7-7 t^{2}} d t=\int_{1}^{\sqrt{2}} \frac{-8 t}{7 t^{2}-16 t+4} d t= \\
= & \int_{1}^{\sqrt{2}} \frac{-8 t}{7\left(t-\frac{2}{7}\right)(t-2)} d t=\int_{1}^{(7 t-2)(t-2)} d t=
\end{aligned}
$$

Decompose the proper rational fraction into elementary fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{-8 t}{(7 t-2)(t-2)}=\frac{A}{7 t-2}+\frac{B}{t-2}=\frac{A(t-2)+B(7 t-2)}{(7 t-2)(t-2)}= \\
& =\frac{A t-2 A+7 B t-2 B}{(7 t-2)(t-2)}=\frac{(A+7 B) t-2(A+B)}{(7 t-2)(t-2)}= \\
& \left\{\begin{array}{l}
A+7 B=-8 \\
-2(A+B)=0
\end{array}\right. \\
& \left\{\begin{array}{l}
A+7 B=-8 \\
A+B=0
\end{array}\right. \\
& \left\{\begin{array}{l}
-B+7 B=-8 \\
A=-B
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-\frac{4}{3} \\
A=\frac{4}{3}
\end{array}\right. \\
& \frac{-8 t}{(7 t-2)(t-2)}=\frac{4}{3(7 t-2)}-\frac{4}{3(t-2)}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =\int_{1}^{\sqrt{2}}\left(\frac{4}{3(7 t-2)}-\frac{4}{3(t-2)}\right) d t=\frac{4}{21} \cdot \int_{1}^{\sqrt{2}} \frac{d(7 t-2)}{7 t-2}-\frac{4}{3} \cdot \int_{1}^{\sqrt{2}} \frac{1}{t-2} d t= \\
& =\frac{4}{21} \cdot \ln \left|7 t-2\left\|_{1}^{\sqrt{2}}-\frac{4}{3} \cdot \ln |t-2|\right\|_{1}^{\sqrt{2}}=\right. \\
& =\frac{4}{21} \cdot \ln |7 \sqrt{2}-2|-\frac{4}{21} \cdot \ln |7 \cdot 1-2|-\frac{4}{3} \cdot \ln |\sqrt{2}-2|+\frac{4}{3} \cdot \ln |1-2|= \\
& =\frac{4}{21} \cdot \ln |7 \sqrt{2}-2|-\frac{4}{21} \cdot \ln 5-\frac{4}{3} \cdot \ln |2-\sqrt{2}|+\frac{4}{3} \cdot \ln 1= \\
& =\frac{4}{21} \cdot \ln \left|\frac{7 \sqrt{2}-2}{5}\right|-\frac{4}{3} \cdot \ln |2-\sqrt{2}|
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-28$ »

Categories: Kuznetsov's Problem Book Integrals Problem 9 | Integrals

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## Problem Kuznetsov Integrals 9-29

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