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

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:

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

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.

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

Explore the detailed step-by-step solution for evaluating the integral $ \int_0^1 \frac{1}{1+\left(\cos\frac{\pi}{2}x\right)}\,dx $. This solution covers trigonometric substitution, important identities, and integration techniques. Ideal for students studying undergraduate calculus.

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