The image contains several integral problems. I will begin by solving a selection of them. Let's start with the integral marked as (4) in problem 31:

$\int_0^\pi \cos^2 x \, dx$

Solution:

We will use a trigonometric identity to simplify this integral. The identity is: $\cos^2 x = \frac{1 + \cos(2x)}{2}$

Thus, the integral becomes: $\int_0^\pi \cos^2 x \, dx = \int_0^\pi \frac{1 + \cos(2x)}{2} \, dx$

We can split this into two separate integrals: $\int_0^\pi \frac{1}{2} \, dx + \int_0^\pi \frac{\cos(2x)}{2} \, dx$

Now, let's solve these integrals:

1. $\int_0^\pi \frac{1}{2} \, dx = \frac{1}{2} \times \pi = \frac{\pi}{2}$
2. $\int_0^\pi \frac{\cos(2x)}{2} \, dx$

For the second part, the integral of $\cos(2x)$ is $\frac{\sin(2x)}{2}$, so: $\int_0^\pi \frac{\cos(2x)}{2} \, dx = \left[\frac{\sin(2x)}{4}\right]_0^\pi = 0$

Thus, the full integral becomes: $\frac{\pi}{2} + 0 = \frac{\pi}{2}$

So, the solution to the integral is:

$\int_0^\pi \cos^2 x \, dx = \frac{\pi}{2}$

Would you like me to solve other integrals from this list or provide further details on this one?

Follow-up questions:

1. How would the identity change for $\sin^2 x$, and how would you apply it to solve a similar integral?
2. Can you compute $\int_0^\pi \cos^2(2x) \, dx$ using the same technique?
3. How would this integral change if the limits were from $0$ to $\frac{\pi}{2}$ instead of $0$ to $\pi$?
4. Could you apply this method to solve $\int_0^{2\pi} \cos^2 x \, dx$?
5. How does this integral compare to integrals involving $\cos(2x)$ directly, rather than squared terms?

Tip:

When solving trigonometric integrals involving squares, always consider using the appropriate double-angle identities to simplify the expression.