Let `x^2 ne npi-1, n in N`, then `intxsqrt((2sin(x^2+1)-sin2(x^2+1))/(2sin(x^2+1)+sin2(x^2+1)))dx` equals

Options

A. `log|1/2sec(x^2+1)|+C`
B. `log|sec(1/2(x^2+1))|+C`
C. `1/2log|sec(x^2+1)|+C`
D. None of these

Answer

To solve the integral

\[ I = \int x \sqrt{\frac{2 \sin(x^2 + 1) - \sin(2(x^2 + 1))}{2 \sin(x^2 + 1) + \sin(2(x^2 + 1))}} \, dx, \]

we will follow these steps:

### Step 1: Substitution
Let \( t = x^2 + 1 \). Then, we differentiate both sides to find \( dt \):
\[ dt = 2x \, dx \implies x \, dx = \frac{1}{2} dt. \]

### Step 2: Rewrite the Integral
Substituting \( t \) into the integral, we have:
\[ I = \int x \sqrt{\frac{2 \sin(t) - \sin(2t)}{2 \sin(t) + \sin(2t)}} \, dx = \int \sqrt{\frac{2 \sin(t) - \sin(2t)}{2 \sin(t) + \sin(2t)}} \cdot \frac{1}{2} dt. \]

### Step 3: Simplify the Expression
Using the identity \( \sin(2t) = 2 \sin(t) \cos(t) \), we can rewrite the integral:
\[ I = \frac{1}{2} \int \sqrt{\frac{2 \sin(t) - 2 \sin(t) \cos(t)}{2 \sin(t) + 2 \sin(t) \cos(t)}} \, dt. \]
Factoring out \( 2 \sin(t) \) from both the numerator and denominator gives:
\[ I = \frac{1}{2} \int \sqrt{\frac{2 \sin(t)(1 - \cos(t))}{2 \sin(t)(1 + \cos(t))}} \, dt = \frac{1}{2} \int \sqrt{\frac{1 - \cos(t)}{1 + \cos(t)}} \, dt. \]

### Step 4: Use Trigonometric Identities
Using the identity \( 1 - \cos(t) = 2 \sin^2\left(\frac{t}{2}\right) \) and \( 1 + \cos(t) = 2 \cos^2\left(\frac{t}{2}\right) \), we can simplify further:
\[ I = \frac{1}{2} \int \sqrt{\frac{2 \sin^2\left(\frac{t}{2}\right)}{2 \cos^2\left(\frac{t}{2}\right)}} \, dt = \frac{1}{2} \int \tan\left(\frac{t}{2}\right) \, dt. \]

### Step 5: Integrate
The integral of \( \tan\left(\frac{t}{2}\right) \) is:
\[ \int \tan\left(\frac{t}{2}\right) \, dt = -\log\left(\cos\left(\frac{t}{2}\right)\right) + C. \]
Thus,
\[ I = \frac{1}{2} \left(-\log\left(\cos\left(\frac{t}{2}\right)\right) + C\right). \]

### Step 6: Back Substitute
Recalling that \( t = x^2 + 1 \), we substitute back:
\[ I = -\frac{1}{2} \log\left(\cos\left(\frac{x^2 + 1}{2}\right)\right) + C. \]

### Final Answer
Thus, the integral evaluates to:
\[ I = -\frac{1}{2} \log\left(\cos\left(\frac{x^2 + 1}{2}\right)\right) + C. \]