Trigonometric integral with cosinus I cannot solve the equation below:
I know what I typed is wrong byt I can't understand where it went sour.
$$\int \frac{1}{\cos(5x)} dx$$
$$\left[
\begin{align*}
5x=s\\
x=\frac{s}{5};
\frac{dx}{ds}=\frac{1}5
\end{align*}
\right]
$$
$$
\frac{1}{5}\int \frac{1}{\cos (s)}\frac{\cos (s)}{\cos (s)}\,ds=
\frac{1}{5}\int \frac{\cos (s)}{\cos^2(s)}\,ds=\frac{1}{5}\int \frac{\cos (s)}{1-\sin^2(s)}\,ds
$$
$$\left[
\begin{align*}
\sin(s)=y\\
\frac{dy}{ds}=-\cos(s)
\end{align*}
\right]
$$
$$
\frac{1}{5} \cdot -\frac{1}{2}\int -\frac{1}{1-y}-\frac{1}{1+y} dy
$$
$$
\left[ -(-\ln(1-y) - \ln(1+y)\right]
=
\left[ \ln\frac{(1-y)}{(1+y)}\right]
=
\left[ \ln\frac{(1-\sin(s))}{(1+\sin(s))}\right]
=
\left[ \ln\frac{(1-\sin(5x))}{(1+\sin(5x))}\right]
$$

Hint. Note that taking the derivative of
$$\ln\left(\frac{1-\sin(5x)}{1+\sin(5x)}\right)=\ln(\underbrace{1-\sin(5x)}_{>0})-\ln(\underbrace{1+\sin(5x)}_{>0})$$
we get
$$\frac{-5\cos(5x)}{1-\sin(5x)}-\frac{5\cos(5x)}{1+\sin(5x)}=
-\frac{10\cos(5x)}{1-\sin^2(5x)}=-\frac{10}{\cos(5x)}.$$
So you are not so far from the correct answer: you just forgot to consider the constant $\frac{1}{5} \cdot -\frac{1}{2}$ in the last line.
