How to solve without involving hyperbolic function. How to solve this integral without involving hyperbolic functions?
$$\int \frac{1}{4-5\sin^2 x}dx$$
The answer is $\frac{1}{4}(\ln (\sin x+2 \cos x)-\ln(2\cos x-\sin x))+c$
 A: \begin{align}I&= \int \frac{\sec^2 x}{4\sec^2 x -5\tan^2 x}\, \mathrm dx \\&= \int\frac{\mathrm du}{4 - u^2} \,\,\,\;\;\; [\textrm{substituting} \;\tan x = u\; \textrm{and using the differential}\; \mathrm du = \sec^2 x\,\mathrm dx \,.]\\& = \frac{1}{4} \ln \left|\frac{2+ u}{2-u}\right| +\rm C\,\,\,\;\;\; \left[\textrm{using the formula} \; \int \frac{1}{a^2 -x^2}\,\mathrm dx= \frac{1}{2a} \ln \left|\frac{a-x}{a+x}\right| \right]\;.\end{align} 
Now replace the value of $u$ and the desired result will come out.
A: \begin{align*}
\int \frac{dx}{4 - 5\sin^2 x} &= \int \frac{dx}{4 \cos^2 x + 4 \sin^2 x - 5 \sin^2 x}\\
&= \int \frac{dx}{4 \cos^2 x - \sin^2 x}\\
&= \int \frac{\sec^2 x}{4 - \tan^2 x} \, dx
\end{align*}
Let $u = \tan x, du = \sec^2 x \, dx$. So
\begin{align*}
\int \frac{dx}{4 - 5\sin^2 x} &= \int \frac{du}{4 - u^2}\\
&= \frac{1}{4} \int \frac{du}{2 - u} + \frac{1}{4} \int \frac{du}{2 + u}\\
&= -\frac{1}{4} \ln |2 - u| + \frac{1}{4} \ln |2 + u| + \cal{C}\\
&= -\frac{1}{4} \ln |2 - \tan x| + \frac{1}{4} \ln |2 + \tan x| + \cal{C}\\
&= -\frac{1}{4} \ln \left |2 - \frac{\sin x}{\cos x} \right | + \frac{1}{4} \ln \left |2 + \frac{\sin x}{\cos x} \right | + \cal{C}\\
&= -\frac{1}{4} \ln \left |\frac{2 \cos x - \sin x}{\cos x} \right | + \frac{1}{4} \ln \left |\frac{2 \cos x + \sin x}{\cos x} \right | + \cal{C}\\
&= -\frac{1}{4} \ln |2 \cos x - \sin x| + \frac{1}{4} \ln |2 \cos x + \sin x| + \cal{C}
\end{align*}
as required to show.
