`inttan^(-1)(sinx/(1+cosx))dx=`

Options

A. `x^(2)/2+c`
B. `x^(2)/3+c`
C. `x^(2)/4+c`
D. `x^(2)/5+c`

Answer

To solve the integral \( \int \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) dx \), we can use trigonometric identities to simplify the expression. Here’s the step-by-step solution:

### Step 1: Simplify the Argument of the Inverse Tangent
We start with the expression inside the inverse tangent:
\[ \frac{\sin x}{1 + \cos x} \]
Using the identity \( 1 + \cos x = 2 \cos^2\left(\frac{x}{2}\right) \) and \( \sin x = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \), we can rewrite:
\[ \frac{\sin x}{1 + \cos x} = \frac{2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right)}{2 \cos^2\left(\frac{x}{2}\right)} = \tan\left(\frac{x}{2}\right) \]
Thus, we have:
\[ \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right) = \frac{x}{2} \]

### Step 2: Substitute Back into the Integral
Now we can substitute this back into the integral:
\[ \int \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) dx = \int \frac{x}{2} dx \]

### Step 3: Integrate
Now we can integrate:
\[ \int \frac{x}{2} dx = \frac{1}{2} \int x \, dx = \frac{1}{2} \cdot \frac{x^2}{2} + C = \frac{x^2}{4} + C \]

### Final Answer
Thus, the final result is:
\[ \int \tan^{-1}\left(\frac{\sin x}{1 + \cos x}\right) dx = \frac{x^2}{4} + C \]

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