`int(cosec theta-cot theta)/(cosectheta+ cot theta)d theta=`

Options

A. `2tan theta/2 - x+c`
B. `2cot theta-theta+c`
C. `2tan theta-theta+c`
D. `2cot theta+theta+c`

Answer

To solve the integral

\[ I = \int \frac{\csc \theta - \cot \theta}{\csc \theta + \cot \theta} \, d\theta, \]

we will simplify the expression step by step.

### Step 1: Rewrite the integral in terms of sine and cosine

Using the definitions of cosecant and cotangent, we have:

\[ \csc \theta = \frac{1}{\sin \theta}, \quad \cot \theta = \frac{\cos \theta}{\sin \theta}. \]

Substituting these into the integral gives:

\[ I = \int \frac{\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta} + \frac{\cos \theta}{\sin \theta}} \, d\theta. \]

### Step 2: Simplify the expression

The expression simplifies to:

\[ I = \int \frac{1 - \cos \theta}{1 + \cos \theta} \, d\theta. \]

### Step 3: Use trigonometric identities

We can use the identity \(1 - \cos \theta = 2 \sin^2 \frac{\theta}{2}\) and \(1 + \cos \theta = 2 \cos^2 \frac{\theta}{2}\):

\[ I = \int \frac{2 \sin^2 \frac{\theta}{2}}{2 \cos^2 \frac{\theta}{2}} \, d\theta = \int \tan^2 \frac{\theta}{2} \, d\theta. \]

### Step 4: Integrate \( \tan^2 \frac{\theta}{2} \)

Recall that:

\[ \tan^2 x = \sec^2 x - 1. \]

Thus,

\[ I = \int \tan^2 \frac{\theta}{2} \, d\theta = \int \left( \sec^2 \frac{\theta}{2} - 1 \right) d\theta = \int \sec^2 \frac{\theta}{2} \, d\theta - \int d\theta. \]

The integral of \( \sec^2 x \) is \( \tan x \), so we get:

\[ I = 2 \tan \frac{\theta}{2} - \theta + C, \]

where \( C \) is the constant of integration.

### Final Answer

Thus, the solution to the integral is:

\[ I = 2 \tan \frac{\theta}{2} - \theta + C. \]

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