`int x sinx sec^(3)x dx` is equal to

Options

A. `(1)/(2)[ sec^(2)x-tanx]+c`
B. `(1)/(2)[x sec^(2)x-tanx]+c`
C. `(1)/(2)[x sec^(2)x+tanx]+c`
D. `(1)/(2)[ sec^(2)x+tanx]+c`

Explanation

To solve the integral \( \int x \sin x \sec^3 x \, dx \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the Integral**:
\[ \int x \sin x \sec^3 x \, dx \]
We can express \( \sec^3 x \) as \( \sec^2 x \cdot \sec x \):
\[ = \int x \sin x \sec^2 x \sec x \, dx \]

2. **Use Trigonometric Identity**:
Recall that \( \sec x = \frac{1}{\cos x} \) and \( \tan x = \frac{\sin x}{\cos x} \). Thus, we can rewrite \( \sin x \sec^2 x \) as:
\[ \sin x \sec^2 x = \tan x \sec x \]
Therefore, the integral becomes:
\[ = \int x \tan x \sec^2 x \, dx \]

3. **Integration by Parts**:
We will use integration by parts where we let:
\[ u = x \quad \text{and} \quad dv = \tan x \sec^2 x \, dx \]
Then, we need to find \( du \) and \( v \):
\[ du = dx \]
To find \( v \), we integrate \( dv \):
\[ v = \int \tan x \sec^2 x \, dx \]
The integral of \( \tan x \sec^2 x \) is \( \frac{1}{2} \tan^2 x \).

4. **Apply Integration by Parts Formula**:
The integration by parts formula is:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ = x \cdot \frac{1}{2} \tan^2 x - \int \frac{1}{2} \tan^2 x \, dx \]

5. **Integrate \( \tan^2 x \)**:
We can express \( \tan^2 x \) using the identity:
\[ \tan^2 x = \sec^2 x - 1 \]
Thus,
\[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \int \sec^2 x \, dx - \int 1 \, dx \]
The integral of \( \sec^2 x \) is \( \tan x \), and the integral of 1 is \( x \):
\[ = \tan x - x \]

6. **Substituting Back**:
Now, substituting back into our integration by parts result:
\[ = x \cdot \frac{1}{2} \tan^2 x - \frac{1}{2} \left( \tan x - x \right) \]
Simplifying:
\[ = \frac{1}{2} x \tan^2 x - \frac{1}{2} \tan x + \frac{1}{2} x + C \]

7. **Final Result**:
Thus, the final result of the integral is:
\[ \int x \sin x \sec^3 x \, dx = \frac{1}{2} x \tan^2 x - \frac{1}{2} \tan x + \frac{1}{2} x + C \]