To evaluate the integral

$\int_0^1 \frac{1}{1 + \left(\cos\frac{\pi}{2}x\right)}\, dx,$

we can proceed as follows:

1. Substitution:
   Let $u = \frac{\pi}{2}x$. Then, $du = \frac{\pi}{2}dx$, and the limits change as follows:

   • When $x = 0$, $u = 0$.
   • When $x = 1$, $u = \frac{\pi}{2}$.

   The integral becomes:

   $\int_0^{\frac{\pi}{2}} \frac{2/\pi}{1 + \cos u}\, du.$

2. Use the identity for $1 + \cos u$: Recall the trigonometric identity: $1 + \cos u = 2\cos^2\left(\frac{u}{2}\right).$

   Therefore, the integral becomes:

   = \frac{1}{\pi} \int_0^{\frac{\pi}{2}} \sec^2\left(\frac{u}{2}\right)\, du.$$

3. Substitute $v = \frac{u}{2}$:
   Let $v = \frac{u}{2}$, so $dv = \frac{1}{2} du$. The limits change to:

   • When $u = 0$, $v = 0$.
   • When $u = \frac{\pi}{2}$, $v = \frac{\pi}{4}$.

   The integral now becomes:

   $\frac{2}{\pi} \int_0^{\frac{\pi}{4}} \sec^2(v)\, dv.$

4. Integrate $\sec^2(v)$: The integral of $\sec^2(v)$ is $\tan(v)$. So:

   $\frac{2}{\pi} \left[ \tan(v) \right]_0^{\frac{\pi}{4}}.$

   Evaluating the limits:

   [ \frac{2}{\pi} \left[ \tan\left(\frac{\pi}{4}\right) - \tan(0) \right] = \frac{2}{\pi} \left[ 1 - 0 \right] = \frac{2}{\pi}. ]

Thus, the value of the integral is:

$\boxed{\frac{2}{\pi}}.$

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How would the integral change if the limits were from 0 to 2 instead of 0 to 1?
2. Can you compute the integral $\int_0^1 \frac{1}{1 + \sin\left(\frac{\pi}{2}x\right)}\, dx$?
3. How does the integral change if the integrand is $\frac{1}{1 + \left(\sin\frac{\pi}{2}x\right)}$ instead of cosine?
4. What if the cosine function is replaced with a tangent function in the integrand?
5. Can you derive the general form for $\int_0^1 \frac{1}{1 + \left(\cos kx\right)}\, dx$?

Tip: When working with trigonometric integrals, consider using known identities to simplify the expression before integrating.