We are asked to solve the integral:

$I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1 + \sin x}$

using the substitution $\sin x$ in terms of $\tan$, but without changing the limits, which we will reapply after resubstitution. Here's how to approach this:

Step 1: Substitution

We use the Weierstrass substitution, which is based on the half-angle identity: $\tan\left(\frac{x}{2}\right) = t$ Thus, the following relations hold: $\sin x = \frac{2t}{1 + t^2}, \quad dx = \frac{2\, dt}{1 + t^2}$

Step 2: Rewrite the integrand

Using these, the integral becomes:

$I = \int \frac{dx}{1 + \sin x} = \int \frac{\frac{2\, dt}{1 + t^2}}{1 + \frac{2t}{1 + t^2}}$

Simplify the denominator: $1 + \frac{2t}{1 + t^2} = \frac{(1 + t^2) + 2t}{1 + t^2} = \frac{1 + 2t + t^2}{1 + t^2}$

Thus, the integral simplifies to:

$I = \int \frac{\frac{2\, dt}{1 + t^2}}{\frac{1 + 2t + t^2}{1 + t^2}} = \int \frac{2\, dt}{1 + 2t + t^2}$

Step 3: Complete the square

We can simplify the denominator by completing the square:

$1 + 2t + t^2 = (t + 1)^2$

So, the integral becomes:

$I = \int \frac{2\, dt}{(t + 1)^2}$

Step 4: Solve the integral

The integral of $\frac{1}{(t + 1)^2}$ is straightforward:

$I = \int 2 \cdot (t + 1)^{-2}\, dt = -\frac{2}{t + 1} + C$

Step 5: Back-substitution

Now, we substitute back $t = \tan\left(\frac{x}{2}\right)$:

$I = -\frac{2}{\tan\left(\frac{x}{2}\right) + 1}$

Step 6: Apply limits

Now, we apply the limits $x = \frac{\pi}{4}$ to $x = \frac{3\pi}{4}$.

• For $x = \frac{\pi}{4}$, we have: $\tan\left(\frac{\pi}{4} \times \frac{1}{2}\right) = \tan\left(\frac{\pi}{8}\right)$

• For ( x = \frac{3\pi