## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\pi / 4} \frac{6 \sin ^{2} x}{3 \cos 2 x-4} d x
$$
## Solution

## Solution №1

We will prove the lemma:

$$
\begin{aligned}
& \int \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}=\frac{1}{a b} \operatorname{arctg}\left(\frac{b}{a} \operatorname{tg} x\right)+C \\
& \int \frac{d x}{a^{2} \cos ^{2} x+b^{2} \sin ^{2} x}=\int \frac{\frac{1}{\cos ^{2} x} d x}{a^{2}+b^{2} \operatorname{tg}^{2} x}=\int \frac{d(\operatorname{tg} x)}{b^{2} \operatorname{tg}^{2} x+a^{2}}=\frac{1}{b^{2}} \int \frac{d(\operatorname{tg} x)}{\operatorname{tg}^{2} x+\frac{a^{2}}{b^{2}}}
\end{aligned}
$$

Let: \( U = \operatorname{tg} x \)

$$
\frac{1}{b^{2}} \int \frac{d(\operatorname{tg} x)}{\operatorname{tg}^{2} x+\frac{a^{2}}{b^{2}}}=\frac{1}{b^{2}} \int \frac{d U}{U^{2}+\left(\frac{a}{b}\right)^{2}}=\frac{1}{b^{2}} \cdot \frac{b}{a} \cdot \operatorname{arctg} \frac{b U}{a}+C=\frac{1}{a b} \operatorname{arctg}\left(\frac{b}{a} \operatorname{tg} x\right)+C \text { Q.E.D. }
$$

Let's proceed to the direct solution:

$$
\begin{aligned}
& \int_{0}^{\frac{\pi}{4}} \frac{6 \sin ^{2} x}{3 \cos 2 x-4} d x=\int_{0}^{\frac{\pi}{4}} \frac{6 \frac{1-\cos 2 x}{2}}{3 \cos 2 x-4} d x=\int_{0}^{\frac{\pi}{4}} \frac{3(1-\cos 2 x)}{3 \cos 2 x-4} d x=-\int_{0}^{\frac{\pi}{4}} \frac{3 \cos (2 x)-4+1}{3 \cos 2 x-4} d x \\
& =-\int_{0}^{\frac{\pi}{4}} d x-\int_{0}^{\frac{\pi}{4}} \frac{1}{3 \cos 2 x-4}=-\int_{0}^{\frac{\pi}{4}} d x-\int_{0}^{\frac{\pi}{4}} \frac{1}{2 \frac{\frac{\pi}{4}+3 \cos 2 x}{2}-7}=-\int_{0}^{\frac{\pi}{4}} d x-\int_{0}^{\frac{\pi}{4}} \frac{1}{6 \cos ^{2} x-7} d x= \\
& =-\int_{0}^{\frac{\pi}{4}} d x-\int_{0}^{\frac{\pi}{4}} \frac{1}{-\cos ^{2} x-7 \sin ^{2} x} d x=-\int_{0}^{\frac{\pi}{4}} d x+\int_{0}^{\frac{\pi}{4}} \frac{d x}{1^{2} \cos ^{2} x+(\sqrt{7})^{2} \sin ^{2} x}
\end{aligned}
$$

Using the lemma, we get:
$$
-\int_{0}^{\frac{\pi}{4}} d x+\int_{0}^{\frac{\pi}{4}} \frac{d x}{1^{2} \cos ^{2} x+(\sqrt{7})^{2} \sin ^{2} x}=-\frac{\pi}{4}+\left.\frac{1}{\sqrt{7}} \operatorname{arctg}(\sqrt{7} \operatorname{tg} x)\right|_{0} ^{\frac{\pi}{4}}=-\frac{\pi}{4}+\frac{1}{\sqrt{7}} \operatorname{arctg} \sqrt{7}
$$

## Solution №2

We will use the substitution:

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

From which:

$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos 2 x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{0}^{\pi / 4} \frac{6 \sin ^{2} x}{3 \cos 2 x-4} d x=\int_{0}^{1} \frac{\frac{6 t^{2}}{1+t^{2}}}{3 \cdot \frac{1-t^{2}}{1+t^{2}}-4} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{1} \frac{\frac{6 t^{2}}{1+t^{2}}}{3-3 t^{2}-4-4 t^{2}} d t= \\
& =\int_{0}^{1} \frac{\frac{6 t^{2}}{1+t^{2}}}{3-3 t^{2}-4-4 t^{2}} d t=-\int_{0}^{1} \frac{6 t^{2}}{\left(1+t^{2}\right)\left(1+7 t^{2}\right)} d t=
\end{aligned}
$$

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

$$
\begin{aligned}
& \frac{6 t^{2}}{\left(1+t^{2}\right)\left(1+7 t^{2}\right)}=\frac{A t+B}{1+t^{2}}+\frac{C t+D}{1+7 t^{2}}= \\
& =\frac{(A t+B) \cdot\left(1+7 t^{2}\right)+(C t+D) \cdot\left(1+t^{2}\right)}{\left(1+t^{2}\right)\left(1+7 t^{2}\right)}= \\
& =\frac{A t+7 A t^{3}+B+7 B t^{2}+C t+C t^{3}+D+D t^{2}}{\left(1+t^{2}\right)\left(1+7 t^{2}\right)}= \\
& =\frac{(7 A+C) t^{3}+(7 B+D) t^{2}+(A+C) t+(B+D)}{\left(1+t^{2}\right)\left(1+7 t^{2}\right)} \\
& \left\{\begin{array}{l}
7 A+C=0 \\
7 B+D=6 \\
A+C=0 \\
B+D=0
\end{array}\right. \\
& \left\{\begin{array}{l}
7 A-A=0 \\
7 B-B=6 \\
C=-A \\
D=-B
\end{array}\right.
\end{aligned}
$$

$$
\begin{aligned}
& \left\{\begin{array}{l}
A=0 \\
6 B=6 \\
C=0 \\
D=-B
\end{array}\right. \\
& \left\{\begin{array}{l}
A=0 \\
B=1 \\
C=0 \\
D=-1
\end{array}\right. \\
& \frac{6 t^{2}}{\left(1+t^{2}\right)\left(1+7 t^{2}\right)}=\frac{1}{1+t^{2}}-\frac{1}{1+7 t^{2}}
\end{aligned}
$$

We get:

$$
\begin{aligned}
& =-\int_{0}^{1}\left(\frac{1}{1+t^{2}}-\frac{1}{1+7 t^{2}}\right) d t=-\int_{0}^{1} \frac{d t}{1+t^{2}}+\frac{1}{7} \cdot \int_{0}^{1} \frac{d t}{t^{2}+\frac{1}{7}}= \\
& =-\left.\operatorname{arctg} t\right|_{0} ^{1}+\left.\frac{1}{7} \cdot \frac{1}{\sqrt{\frac{1}{7}}} \cdot \operatorname{arctg} \frac{t}{\sqrt{\frac{1}{7}}}\right|_{0} ^{1}= \\
& =-\operatorname{arctg} 1+\operatorname{arctg} 0+\frac{1}{\sqrt{7}} \cdot \operatorname{arctg}(\sqrt{7} \cdot 1)-\frac{1}{\sqrt{7}} \cdot \operatorname{arctg}(\sqrt{7} \cdot 0)= \\
& =-\frac{\pi}{4}+0+\frac{1}{\sqrt{7}} \cdot \operatorname{arctg} \sqrt{7}-0=-\frac{\pi}{4}+\frac{1}{\sqrt{7}} \cdot \operatorname{arctg} \sqrt{7}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5 $\% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-12$ »

