Evaluate the following integrals.
`int cosx tan^(3) x dx`

Answer

To evaluate the integral \( \int \cos x \tan^3 x \, dx \), we can follow these steps:

### Step 1: Rewrite \( \tan^3 x \)
We know that \( \tan x = \frac{\sin x}{\cos x} \). Therefore, we can express \( \tan^3 x \) as:
\[ \tan^3 x = \left(\frac{\sin x}{\cos x}\right)^3 = \frac{\sin^3 x}{\cos^3 x} \]

### Step 2: Substitute into the integral
Substituting this into the integral gives:
\[ \int \cos x \tan^3 x \, dx = \int \cos x \cdot \frac{\sin^3 x}{\cos^3 x} \, dx = \int \frac{\sin^3 x}{\cos^2 x} \, dx \]

### Step 3: Rewrite \( \sin^3 x \)
We can express \( \sin^3 x \) in terms of \( \cos x \):
\[ \sin^3 x = \sin x (1 - \cos^2 x) \]
Thus, we can rewrite the integral as:
\[ \int \frac{\sin^3 x}{\cos^2 x} \, dx = \int \frac{\sin x (1 - \cos^2 x)}{\cos^2 x} \, dx = \int \left(\frac{\sin x}{\cos^2 x} - \frac{\sin x \cos^2 x}{\cos^2 x}\right) \, dx \]
This simplifies to:
\[ \int \left(\frac{\sin x}{\cos^2 x} - \sin x\right) \, dx \]

### Step 4: Split the integral
Now we can split the integral into two parts:
\[ \int \frac{\sin x}{\cos^2 x} \, dx - \int \sin x \, dx \]

### Step 5: Solve the first integral
Let \( u = \cos x \), then \( du = -\sin x \, dx \) or \( \sin x \, dx = -du \). The first integral becomes:
\[ \int \frac{\sin x}{\cos^2 x} \, dx = -\int \frac{1}{u^2} \, du = \frac{1}{u} = \frac{1}{\cos x} \]

### Step 6: Solve the second integral
The second integral is straightforward:
\[ \int \sin x \, dx = -\cos x \]

### Step 7: Combine the results
Combining both parts, we have:
\[ \int \cos x \tan^3 x \, dx = \frac{1}{\cos x} + \cos x + C \]
where \( C \) is the constant of integration.

### Final Answer
Thus, the final result is:
\[ \int \cos x \tan^3 x \, dx = \sec x + \cos x + C \]