Evaluate: (i) `int1/(x^2)cos^2(1/x)\ dx` (ii) `intsec^4xtanx\ dx`

Answer

Let's evaluate the integrals step by step.

### (i) Evaluate the integral \( I_1 = \int \frac{1}{x^2} \cos^2\left(\frac{1}{x}\right) \, dx \)

**Step 1: Substitution**
Let \( t = \frac{1}{x} \). Then, \( x = \frac{1}{t} \) and differentiating both sides gives:
\[ dx = -\frac{1}{t^2} \, dt \]
Also, \( \frac{1}{x^2} = t^2 \).

**Step 2: Rewrite the integral**
Substituting these into the integral, we get:
\[ I_1 = \int t^2 \cos^2(t) \left(-\frac{1}{t^2}\right) dt = -\int \cos^2(t) \, dt \]

**Step 3: Use the trigonometric identity**
Using the identity \( \cos^2(t) = \frac{1 + \cos(2t)}{2} \):
\[ I_1 = -\int \frac{1 + \cos(2t)}{2} \, dt = -\frac{1}{2} \int (1 + \cos(2t)) \, dt \]

**Step 4: Separate the integral**
\[ I_1 = -\frac{1}{2} \left( \int 1 \, dt + \int \cos(2t) \, dt \right) \]

**Step 5: Evaluate the integrals**
The first integral is:
\[ \int 1 \, dt = t \]
The second integral is:
\[ \int \cos(2t) \, dt = \frac{1}{2} \sin(2t) \]
Thus, we have:
\[ I_1 = -\frac{1}{2} \left( t + \frac{1}{2} \sin(2t) \right) + C \]

**Step 6: Substitute back for \( t \)**
Since \( t = \frac{1}{x} \):
\[ I_1 = -\frac{1}{2} \left( \frac{1}{x} + \frac{1}{2} \sin\left(\frac{2}{x}\right) \right) + C \]
Final result:
\[ I_1 = -\frac{1}{2x} - \frac{1}{4} \sin\left(\frac{2}{x}\right) + C \]