## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\arcsin \sqrt{7 / 8}} \frac{6 \sin ^{2} x}{4+3 \cos 2 x} d x
$$
## Solution

We will use the substitution:

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

From which:

$$
\begin{aligned}
& \sin ^{2} x=\frac{t^{2}}{1+t^{2}}, \cos 2 x=\frac{1-t^{2}}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=0 \Rightarrow t=\operatorname{tg} 0=0 \\
& x=\arcsin \sqrt{\frac{7}{8}} \Rightarrow t=\operatorname{tg}\left(\arcsin \sqrt{\frac{7}{8}}\right)=\sqrt{\frac{1}{1-\sin ^{2}\left(\arcsin \sqrt{\frac{7}{8}}\right)}-1}= \\
& =\sqrt{\frac{1}{1-\frac{7}{8}}-1}=\sqrt{8-1}=\sqrt{7}
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
\arcsin \sqrt{7 / 8} & \frac{6 \sin ^{2} x}{4+3 \cos 2 x} d x=\int_{0}^{\sqrt{7}} \frac{\frac{6 t^{2}}{1+t^{2}}}{4+3 \cdot \frac{1-t^{2}}{1+t^{2}}} \cdot \frac{d t}{1+t^{2}}= \\
= & \int_{0}^{\sqrt{7}} \frac{6 t^{2}}{4+4 t^{2}+3-3 t^{2}} \cdot \frac{d t}{1+t^{2}}=\int_{0}^{\sqrt{7}} \frac{6 t^{2}}{\left(t^{2}+7\right)\left(1+t^{2}\right)} d t=
\end{aligned}
$$

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

$$
\frac{6 t^{2}}{\left(t^{2}+7\right)\left(1+t^{2}\right)}=\frac{A t+B}{7+t^{2}}+\frac{C t+D}{1+t^{2}}=
$$

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

We get:

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

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}+9-22 »

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

- Last edited: 16:36, 13 May 2009.
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## Problem Kuznetsov Integrals 9-23

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