## Condition of the problem

Calculate the definite integral:

$$
\int_{\pi / 4}^{\arcsin (2 / \sqrt{5})} \frac{4 \operatorname{tg} x-5}{4 \cos ^{2} x-\sin 2 x+1} d x
$$
## Solution

We will use the substitution:

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

From which:

$$
\begin{aligned}
& \sin 2 x=\frac{2 t}{1+t^{2}}, \cos ^{2} x=\frac{1}{1+t^{2}}, d x=\frac{d t}{1+t^{2}} \\
& x=\frac{\pi}{4} \Rightarrow t=\operatorname{tg} \frac{\pi}{4}=1 \\
& x=\arcsin \frac{2}{\sqrt{5}} \Rightarrow t=\operatorname{tg}\left(\arcsin \frac{2}{\sqrt{5}}\right)=\sqrt{\frac{1}{1-\sin ^{2}\left(\arcsin \frac{2}{\sqrt{5}}\right)}-1}= \\
& =\sqrt{\frac{1}{1-\frac{4}{5}}-1}=\sqrt{5-1}=\sqrt{4}=2
\end{aligned}
$$

Substitute:

$$
\begin{aligned}
& \int_{\pi / 4}^{\arcsin (2 / \sqrt{5})} \frac{4 \operatorname{tg} x-5}{4 \cos ^{2} x-\sin 2 x+1} d x=\int_{1}^{2} \frac{4 t-5}{\frac{4}{1+t^{2}}-\frac{2 t}{1+t^{2}}+1} \cdot \frac{d t}{1+t^{2}}= \\
= & \int_{1}^{2} \frac{4 t-5}{4-2 t+1+t^{2}} d t=\int_{1}^{2} \frac{2(2 t-2)-1}{t^{2}-2 t+5} d t=2 \int_{1}^{2} \frac{d\left(t^{2}-2 t+5\right)}{t^{2}-2 t+5} d t-\int_{1}^{2} \frac{1}{(t-1)^{2}+4} d t= \\
= & \left.2 \ln \left(t^{2}-2 t+5\right)\right|_{1} ^{2}-\left.\frac{1}{2} \operatorname{arctg} \frac{t-1}{2}\right|_{1} ^{2}= \\
= & 2 \ln \left(2^{2}-2 \cdot 2+5\right)-2 \ln \left(1^{2}-2 \cdot 1+5\right)-\frac{1}{2} \operatorname{arctg} \frac{2-1}{2}+\frac{1}{2} \operatorname{arctg} \frac{1-1}{2}= \\
= & 2 \ln 5-2 \ln 4-\frac{1}{2} \operatorname{arctg} \frac{1}{2}+\operatorname{arctg} 0=2 \ln \frac{5}{4}-\frac{1}{2} \operatorname{arctg} \frac{1}{2}
\end{aligned}
$$

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\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+9-21 »

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

- Last modified: 12:53, 13 May 2009.
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## Problem Kuznetsov Integrals 9-22

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