Integrate Tan/Sin Without Special Methods

[tandoorichicken]

How do you do the following integral without using any special methods, i.e., integration by parts, etc.

$$\int \frac{\tan x}{\sin x} \,dx$$

?

 

[BLaH!]

What does $$\tan x$$ equal? Hint: $$\tan x = \frac{\sin x}{?}$$ You should remember what the "?" is!

Anyway, after you make that simplification you will have $$\int \sec x \ dx$$

Unfortunately this integral is tricky to evaluate and DOES require some special tricks to solve (not integration by parts though). Check out this website to see how: integral of secant

 

[wais]

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.