Let the definite integral $ I_n\equal{}\int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^n}\ (n\equal{}0,\ \pm 1,\ \pm 2,\ \cdots )$.

(1) Find $ I_0,\ I_{\minus{}1},\ I_2$.

(2) Find $ I_1$.

(3) Express $ I_{n\plus{}2}$ in terms of $ I_n$.

(4) Find $ I_{\minus{}3},\ I_{\minus{}2},\ I_3$.

(5) Evaluate the definite integrals $ \int_0^1 \sqrt{x^2\plus{}1}\ dx,\ \int_0^1 \frac{dx}{(x^2\plus{}1)^2}\ dx$ in using the avobe results.

You are not allowed to use the formula of integral for $ \sqrt{x^2\plus{}1}$ directively here.
Let's solve the problem step-by-step.

### Part (1): Find \( I_0, I_{-1}, I_2 \)

1. **Finding \( I_0 \):**
   \[
   I_0 = \int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^0} = \int_0^{\frac{\pi}{4}} 1 \, dx = \left. x \right|_0^{\frac{\pi}{4}} = \frac{\pi}{4}
   \]

2. **Finding \( I_{-1} \):**
   \[
   I_{-1} = \int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^{-1}} = \int_0^{\frac{\pi}{4}} \cos x \, dx
   \]
   Using the antiderivative of \(\cos x\):
   \[
   \int \cos x \, dx = \sin x
   \]
   Therefore,
   \[
   I_{-1} = \left. \sin x \right|_0^{\frac{\pi}{4}} = \sin \frac{\pi}{4} - \sin 0 = \frac{\sqrt{2}}{2} - 0 = \frac{\sqrt{2}}{2}
   \]

3. **Finding \( I_2 \):**
   \[
   I_2 = \int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^2} = \int_0^{\frac{\pi}{4}} \sec^2 x \, dx
   \]
   Using the antiderivative of \(\sec^2 x\):
   \[
   \int \sec^2 x \, dx = \tan x
   \]
   Therefore,
   \[
   I_2 = \left. \tan x \right|_0^{\frac{\pi}{4}} = \tan \frac{\pi}{4} - \tan 0 = 1 - 0 = 1
   \]

### Part (2): Find \( I_1 \)

1. **Finding \( I_1 \):**
   \[
   I_1 = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos x} = \int_0^{\frac{\pi}{4}} \sec x \, dx
   \]
   Using the antiderivative of \(\sec x\):
   \[
   \int \sec x \, dx = \ln |\sec x + \tan x|
   \]
   Therefore,
   \[
   I_1 = \left. \ln |\sec x + \tan x| \right|_0^{\frac{\pi}{4}} = \ln (\sec \frac{\pi}{4} + \tan \frac{\pi}{4}) - \ln (\sec 0 + \tan 0)
   \]
   Since \(\sec \frac{\pi}{4} = \sqrt{2}\) and \(\tan \frac{\pi}{4} = 1\):
   \[
   I_1 = \ln (\sqrt{2} + 1) - \ln 1 = \ln (\sqrt{2} + 1)
   \]

### Part (3): Express \( I_{n+2} \) in terms of \( I_n \)

1. **Expressing \( I_{n+2} \) in terms of \( I_n \):**
   Using integration by parts, let \( u = \sec^{n-1} x \) and \( dv = \sec x \, dx \):
   \[
   I_n = \int_0^{\frac{\pi}{4}} \sec^n x \, dx
   \]
   \[
   I_{n+2} = \int_0^{\frac{\pi}{4}} \sec^{n+2} x \, dx
   \]
   Using the reduction formula for \(\sec^n x\):
   \[
   I_{n+2} = \frac{n+1}{n} I_n
   \]

### Part (4): Find \( I_{-3}, I_{-2}, I_3 \)

1. **Finding \( I_{-3} \):**
   \[
   I_{-3} = \int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^{-3}} = \int_0^{\frac{\pi}{4}} \cos^3 x \, dx
   \]
   Using the reduction formula for \(\cos^n x\):
   \[
   I_{-3} = \frac{2}{3} I_{-1} = \frac{2}{3} \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{3}
   \]

2. **Finding \( I_{-2} \):**
   \[
   I_{-2} = \int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^{-2}} = \int_0^{\frac{\pi}{4}} \cos^2 x \, dx
   \]
   Using the reduction formula for \(\cos^n x\):
   \[
   I_{-2} = \frac{1}{2} I_0 = \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{8}
   \]

3. **Finding \( I_3 \):**
   \[
   I_3 = \int_0^{\frac{\pi}{4}} \frac{dx}{(\cos x)^3} = \int_0^{\frac{\pi}{4}} \sec^3 x \, dx
   \]
   Using the reduction formula for \(\sec^n x\):
   \[
   I_3 = \frac{2}{1} I_1 = 2 \cdot \ln (\sqrt{2} + 1) = 2 \ln (\sqrt{2} + 1)
   \]

### Part (5): Evaluate the definite integrals

1. **Evaluate \( \int_0^1 \sqrt{x^2 + 1} \, dx \):**
   Using the substitution \( x = \tan \theta \), \( dx = \sec^2 \theta \, d\theta \):
   \[
   \int_0^1 \sqrt{x^2 + 1} \, dx = \int_0^{\frac{\pi}{4}} \sec^3 \theta \, d\theta = I_3 = 2 \ln (\sqrt{2} + 1)
   \]

2. **Evaluate \( \int_0^1 \frac{dx}{(x^2 + 1)^2} \):**
   Using the substitution \( x = \tan \theta \), \( dx = \sec^2 \theta \, d\theta \):
   \[
   \int_0^1 \frac{dx}{(x^2 + 1)^2} = \int_0^{\frac{\pi}{4}} \cos^2 \theta \, d\theta = I_{-2} = \frac{\pi}{8}
   \]

The final answer is \(\boxed{\frac{2 + \pi}{8}}\)