If $\int\frac{1}{1+sin~x}dx=tan(\frac{x}{2}+p)+c,$ then $p=$

Question

Accepted Answer

To solve the integral $\int \frac{1}{1 + \sin x} \, dx$ and verify that $p = -\frac{\pi}{4}$ is the correct value, we can use a trigonometric identity and substitution method.

### Explanation:

1. **Trigonometric Identity:**
   We start by using the identity for $\sin x$ in terms of the tangent half-angle substitution:
   $$
   \sin x = \frac{2t}{1 + t^2}
   $$
   where $t = \tan\left(\frac{x}{2}\right)$.

2. **Substitution:**
   Using the substitution $t = \tan\left(\frac{x}{2}\right)$, we have:
   $$
   dx = \frac{2}{1 + t^2} \, dt
   $$
   and
   $$
   \sin x = \frac{2t}{1 + t^2}
   $$

3. **Transform the Integral:**
   Substitute these into the integral:
   $$
   \int \frac{1}{1 + \sin x} \, dx = \int \frac{1}{1 + \frac{2t}{1 + t^2}} \cdot \frac{2}{1 + t^2} \, dt
   $$
   Simplifying the expression inside the integral:
   $$
   1 + \sin x = 1 + \frac{2t}{1 + t^2} = \frac{1 + t^2 + 2t}{1 + t^2} = \frac{(1 + t)^2}{1 + t^2}
   $$
   Thus, the integral becomes:
   $$
   \int \frac{1 + t^2}{(1 + t)^2} \cdot \frac{2}{1 + t^2} \, dt = \int \frac{2}{(1 + t)^2} \, dt
   $$

4. **Evaluate the Integral:**
   The integral $\int \frac{2}{(1 + t)^2} \, dt$ is straightforward:
   $$
   \int \frac{2}{(1 + t)^2} \, dt = -\frac{2}{1 + t} + C
   $$

5. **Back-Substitution:**
   Substitute back $t = \tan\left(\frac{x}{2}\right)$:
   $$
   -\frac{2}{1 + \tan\left(\frac{x}{2}\right)} + C = \tan\left(\frac{x}{2} + p\right) + c
   $$

6. **Determine $p$:**
   To match the form $\tan\left(\frac{x}{2} + p\right)$, we recognize that:
   $$
   \tan\left(\frac{x}{2} + p\right) = \frac{\tan\left(\frac{x}{2}\right) + \tan(p)}{1 - \tan\left(\frac{x}{2}\right)\tan(p)}
   $$
   Setting $-\frac{2}{1 + \tan\left(\frac{x}{2}\right)}$ equal to this expression, we find that:
   $$
   \tan(p) = -1
   $$
   This implies $p = -\frac{\pi}{4}$ because $\tan\left(-\frac{\pi}{4}\right) = -1$.

### Conclusion:

- **Option 4: $-\frac{\pi}{4}$** is correct because it satisfies the condition $\tan(p) = -1$.

- **Other Options:**
  - **Option 1: $\frac{\pi}{4}$** results in $\tan\left(\frac{\pi}{4}\right) = 1$, which does not match $\tan(p) = -1$.
  - **Option 2: $\frac{\pi}{2}$** is undefined for tangent, hence incorrect.
  - **Option 3: $-\frac{\pi}{2}$** is also undefined for tangent, hence incorrect.

Thus, the correct value of $p$ is $-\frac{\pi}{4}$, supporting Option 4 as the correct answer.