To integrate this integral without using any special methods, we can use the substitution method. Let u = sin x, then du = cos x dx. We can rewrite the integral as:
\int \frac{\tan x}{\sin x} \,dx = \int \frac{\tan x}{u} \,du
Next, we can use the trigonometric identity tan x = sin x / cos x to simplify the integrand:
\int \frac{\tan x}{u} \,du = \int \frac{\sin x}{u \cos x} \,du
Using the substitution u = sin x, we can also rewrite cos x as \sqrt{1-u^2}:
\int \frac{\sin x}{u \cos x} \,du = \int \frac{1}{u \sqrt{1-u^2}} \,du
Now, we can use the substitution v = 1-u^2, then dv = -2u du:
\int \frac{1}{u \sqrt{1-u^2}} \,du = -\frac{1}{2} \int \frac{1}{\sqrt{v}} \,dv
Solving this integral, we get:
-\frac{1}{2} \int \frac{1}{\sqrt{v}} \,dv = -\frac{1}{2} \sqrt{v} + C = -\frac{1}{2} \sqrt{1-u^2} + C
Substituting back u = sin x, we get the final answer of:
\int \frac{\tan x}{\sin x} \,dx = -\frac{1}{2} \sqrt{1-\sin^2 x} + C = -\frac{1}{2} \sqrt{1-\frac{\sin^2 x}{1}} + C = -\frac{1}{2} \sqrt{1-\cos^2 x} + C = -\frac{1}{2} \sin x + C
Therefore, we have successfully integrated \int \frac{\tan x}{\sin x} \,dx without using any special methods. This approach may require some extra steps and algebraic manipulation, but it is a useful method to integrate trigonometric functions without relying on integration by parts or other special methods.
