`int(sin3x)/(cosx)dx=`

Options

A. `log|cosx|+2cos^2x+c`
B. `log|secx|+2cos^2x+c`
C. `log|cosx|-2cos^2x+c`
D. `log|secx|-2cos^2x+c`

Answer

To solve the integral \( I = \int \frac{\sin 3x}{\cos x} \, dx \), we can follow these steps:

### Step 1: Rewrite \(\sin 3x\)
We can use the trigonometric identity for \(\sin 3x\):
\[ \sin 3x = 3 \sin x - 4 \sin^3 x \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{3 \sin x - 4 \sin^3 x}{\cos x} \, dx \]

### Step 2: Separate the Integral
We can separate the integral into two parts:
\[ I = \int \frac{3 \sin x}{\cos x} \, dx - 4 \int \frac{\sin^3 x}{\cos x} \, dx \]
This can be simplified to:
\[ I = 3 \int \tan x \, dx - 4 \int \frac{\sin^3 x}{\cos x} \, dx \]

### Step 3: Integrate the First Part
The integral of \(\tan x\) is:
\[ \int \tan x \, dx = -\log |\cos x| + C \]
Thus, we have:
\[ 3 \int \tan x \, dx = -3 \log |\cos x| \]

### Step 4: Simplify the Second Integral
Now we need to simplify the second integral:
\[ \int \frac{\sin^3 x}{\cos x} \, dx \]
We can rewrite \(\sin^3 x\) as \(\sin x (1 - \cos^2 x)\):
\[ \int \frac{\sin^3 x}{\cos x} \, dx = \int \frac{\sin x (1 - \cos^2 x)}{\cos x} \, dx = \int \frac{\sin x}{\cos x} \, dx - \int \frac{\sin x \cos^2 x}{\cos x} \, dx \]
This gives us:
\[ \int \tan x \, dx - \int \cos x \, dx \]

### Step 5: Change of Variables
For the integral \(\int \sin x \, dx\), we can use the substitution \(t = \cos x\), which gives \(dt = -\sin x \, dx\). Thus:
\[ \int \sin x \, dx = -\int dt = -t + C = -\cos x + C \]

### Step 6: Combine the Results
Now we can combine everything:
\[ I = -3 \log |\cos x| - 4 \left( \int \tan x \, dx - \left(-\cos x\right) \right) \]
This gives us:
\[ I = -3 \log |\cos x| - 4 \left( -\log |\cos x| + \cos x \right) \]
Simplifying this:
\[ I = -3 \log |\cos x| + 4 \log |\cos x| - 4 \cos x \]
Thus:
\[ I = \log |\cos x| - 4 \cos x + C \]

### Final Answer
The final result is:
\[ I = \log |\cos x| - 4 \cos x + C \]