The integral `I=int((1)/(x.secx)-ln(x^(sinx))dx` simplifies to (where, c is the constant of integration)

Options

A. `(ln x)(sinx)+c`
B. `(lnx)(cosx)+c`
C. `xln(sinx)+c`
D. `xln(cosx)+c`

Answer

To solve the integral \( I = \int \left( \frac{1}{x \sec x} - \ln(x^{\sin x}) \right) dx \), we will follow these steps:

### Step 1: Simplify the logarithmic term
We start by simplifying the logarithmic term:
\[ \ln(x^{\sin x}) = \sin x \cdot \ln x \]
Thus, we can rewrite the integral as:
\[ I = \int \left( \frac{1}{x \sec x} - \sin x \ln x \right) dx \]

### Step 2: Separate the integral
Now, we can separate the integral into two parts:
\[ I = \int \frac{1}{x \sec x} \, dx - \int \sin x \ln x \, dx \]

### Step 3: Simplify the first integral
The first integral can be simplified:
\[ \frac{1}{x \sec x} = \frac{\cos x}{x} \]
Thus, we have:
\[ I = \int \frac{\cos x}{x} \, dx - \int \sin x \ln x \, dx \]

### Step 4: Solve the second integral using integration by parts
For the second integral \( \int \sin x \ln x \, dx \), we will use integration by parts. Let:
- \( u = \ln x \) (hence \( du = \frac{1}{x} dx \))
- \( dv = \sin x \, dx \) (hence \( v = -\cos x \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int \sin x \ln x \, dx = -\ln x \cos x - \int -\cos x \cdot \frac{1}{x} \, dx \]
This simplifies to:
\[ \int \sin x \ln x \, dx = -\ln x \cos x + \int \frac{\cos x}{x} \, dx \]

### Step 5: Substitute back into the integral
Now substituting back into our expression for \( I \):
\[ I = \int \frac{\cos x}{x} \, dx - \left( -\ln x \cos x + \int \frac{\cos x}{x} \, dx \right) \]
This simplifies to:
\[ I = \int \frac{\cos x}{x} \, dx + \ln x \cos x - \int \frac{\cos x}{x} \, dx \]
The \( \int \frac{\cos x}{x} \, dx \) terms cancel out:
\[ I = \ln x \cos x + C \]

### Final Answer
Thus, the integral simplifies to:
\[ I = \ln x \cos x + C \]