`int(sinx)/(1+cos^2x)dx=`

Options

A. `2cot^(-1)(cosx)+c`
B. `2tan^(-1)(cosx)+c`
C. `cot^(-1)(cosx)+c`
D. `tan^(-1)(cosx)+c`

Answer

To solve the integral \(\int \frac{\sin x}{1 + \cos^2 x} \, dx\), we can follow these steps:

### Step 1: Substitution
Let \( t = \cos x \). Then, the derivative of \( t \) with respect to \( x \) is:
\[ dt = -\sin x \, dx \quad \Rightarrow \quad \sin x \, dx = -dt \]
Now, we can rewrite the integral in terms of \( t \):
\[ \int \frac{\sin x}{1 + \cos^2 x} \, dx = \int \frac{-dt}{1 + t^2} \]

### Step 2: Simplifying the Integral
The integral now becomes:
\[ -\int \frac{dt}{1 + t^2} \]
We recognize that the integral \(\int \frac{dt}{1 + t^2}\) is a standard integral that equals \(\tan^{-1}(t)\).

### Step 3: Evaluating the Integral
Thus, we have:
\[ -\int \frac{dt}{1 + t^2} = -\tan^{-1}(t) + C \]
Substituting back \( t = \cos x \):
\[ -\tan^{-1}(\cos x) + C \]

### Step 4: Final Answer
The final answer is:
\[ \int \frac{\sin x}{1 + \cos^2 x} \, dx = -\tan^{-1}(\cos x) + C \]