## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2 \operatorname{arctan} \frac{1}{3}} \frac{\cos x d x}{(1-\sin x)(1+\cos x)}
$$
## Solution

We will use the universal substitution:

$$
t=\operatorname{tg} \frac{x}{2}
$$

From which:

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

Substitute:

$$
\begin{aligned}
& \int_{0}^{2 \operatorname{arctg} \frac{1}{3}} \frac{\cos x d x}{(1-\sin x)(1+\cos x)}=\int_{0}^{\frac{1}{3}} \frac{\frac{1-t^{2}}{1+t^{2}} \cdot \frac{2 d t}{1+t^{2}}}{\left(1-\frac{2 t}{1+t^{2}}\right)\left(1+\frac{1-t^{2}}{1+t^{2}}\right)}= \\
& =\int_{0}^{\frac{1}{3}} \frac{2\left(1-t^{2}\right) d t}{\left(1+t^{2}-2 t\right)\left(1+t^{2}+1-t^{2}\right)}=\int_{0}^{\frac{1}{3}} \frac{2\left(1-t^{2}\right) d t}{(1-t)^{2} \cdot 2}=\int_{0}^{\frac{1}{3}} \frac{(1-t)(1+t)}{(1-t)^{2}} d t= \\
& =-\int_{0}^{\frac{1}{3}} \frac{t-1+2}{t-1} d t=-\int_{0}^{\frac{1}{3}} d t-2 \int_{0}^{\frac{1}{3}} \frac{1}{t-1} d t=-\left.t\right|_{0} ^{\frac{1}{3}}-2 \ln |t-1|_{0}^{\frac{1}{3}}= \\
& =-\frac{1}{3}+0-2 \ln \left|\frac{1}{3}-1\right|+2 \ln |0-1|=-\frac{1}{3}-2 \ln \frac{2}{3}
\end{aligned}
$$

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%D1%82%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_8-16»

Categories: Kuznetsov's Problem Book Integrals Problem 8 | Integrals

- Last edited: 19:37, 27 May 2009.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 8-17

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