How do you find the antiderivative of #(sinx - cosx)/cosx dx#?

Question

Cameron Alexander · Accepted Answer

Hello,

The answer is #-ln(|cos(x)|) - x + c#.

Write #\frac{\sin(x)-\cos(x)}{\cos(x)} = \frac{\sin(x)}{\cos(x)} - 1 = -\frac{\cos'(x)}{cos(x)} -1#.

You recognize #\int (u')/u = ln(|u|) + c#, therefore the antiderivative of your function is

#-ln(|cos(x)|) - x + c#

where c is an arbitrary real constant.

Cameron Alexander · Answer

undefined To find the antiderivative of $ \frac{\sin x - \cos x}{\cos x} $ with respect to $ x $:

1. Split the expression into two separate fractions: $ \frac{\sin x}{\cos x} - \frac{\cos x}{\cos x} $.

2. Rewrite $ \frac{\sin x}{\cos x} $ as $ \tan x $, since $ \tan x = \frac{\sin x}{\cos x} $.

3. Rewrite $ \frac{\cos x}{\cos x} $ as $ 1 $.

4. Now, the integral becomes $ \tan x - 1 $.

5. Integrate $ \tan x $ and $ 1 $ separately.

6. The antiderivative of $ \tan x $ is $ -\ln|\cos x| + C $, where $ C $ is the constant of integration.

7. The antiderivative of $ 1 $ is $ x $.

8. Combine the results to get the final antiderivative.

Therefore, the antiderivative of $ \frac{\sin x - \cos x}{\cos x} $ with respect to $ x $ is $ -\ln|\cos x| + x + C $, where $ C $ is the constant of integration.