7. Evaluate the indefinite integral. (Use \( C \) for the constant of integration.) \[ \int \sin (t) \sqrt{1+\cos (t)} d t \]

To evaluate the integral \[ \int \sin(t) \sqrt{1+\cos(t)} \, dt, \] we can use a substitution. Let \( u = 1 + \cos(t) \). Then, the derivative \( du = -\sin(t) \, dt \), or \( dt = -\frac{du}{\sin(t)} \). Now, we can rearrange our integral: \[ \sin(t) \, dt = -du \quad \text{implying} \quad dt = -\frac{du}{\sin(t)}. \] However, this approach can become a bit complicated, so let’s instead look for a simpler representation. Notice that: \[ \sqrt{1+\cos(t)} = \sqrt{2} \cos\left(\frac{t}{2}\right) \quad \text{because} \quad 1+\cos(t) = 2\cos^2\left(\frac{t}{2}\right). \] Thus we can rewrite our integral as: \[ \int \sin(t) \sqrt{2} \cos\left(\frac{t}{2}\right) \, dt = \sqrt{2} \int \sin(t) \cos\left(\frac{t}{2}\right) \, dt. \] To integrate \( \sin(t) \cos\left(\frac{t}{2}\right) \), we can use the product-to-sum identities: \[ \sin(t) \cos\left(\frac{t}{2}\right) = \frac{1}{2} \left( \sin\left(t+\frac{t}{2}\right) + \sin\left(t-\frac{t}{2}\right)\right). \] Simplifying gives: \[ \sin(t) \cos\left(\frac{t}{2}\right) = \frac{1}{2} \left( \sin\left(\frac{3t}{2}\right) + \sin\left(\frac{t}{2}\right)\right). \] Now, we can integrate: \[ \int \sin(t) \cos\left(\frac{t}{2}\right) \, dt = \frac{1}{2} \left(-\frac{2}{3} \cos\left(\frac{3t}{2}\right) - 2 \cos\left(\frac{t}{2}\right)\right) + C. \] Putting back our constant factor of \(\sqrt{2}\): \[ \int \sin(t) \sqrt{1+\cos(t)} \, dt = -\frac{\sqrt{2}}{3} \cos\left(\frac{3t}{2}\right) - \sqrt{2} \cos\left(\frac{t}{2}\right) + C. \] Thus, the final answer is: \[ -\frac{\sqrt{2}}{3} \cos\left(\frac{3t}{2}\right) - \sqrt{2} \cos\left(\frac{t}{2}\right) + C. \]