Question

Match List-I with List-II.

List-I (Complex integrate)	List-II (Value)
A. \( \displaystyle \int_{0}^{2\pi} \cos^2 x \, dx \) B. \( \displaystyle \int_{0}^{\pi} \cos x \, dx \) C. \( \displaystyle \oint_C dz \), where \(C\) is a simple closed curve D. \( \displaystyle \oint_C \frac{dz}{z^2-1} \), around circle \(|z-1|=1\)	I. \(4\pi i\) II. \(0\) III. \(\frac{\pi}{4}\) IV. \( \frac{1}{4}\cosh 2 - \frac{1}{2}\sinh 2 + \frac{1}{2}\pi i \sinh 2 \)

• A. A-I, B-II, C-III, D-IV
• B. A-III, B-IV, C-II, D-I
• C. A-III, B-IV, C-I, D-II
• D. A-IV, B-III, C-II, D-I

Hint:

Closed contour of analytic function → zero; residues give non-zero integrals.

Answer 1

The Correct Option is B

Concept: We use standard definite integrals and complex integration theorems:
• Real integrals
• Cauchy's integral theorem
• Residue theorem

Step 1: Evaluate A.
\[ \int_0^{2\pi} \cos^2 x \, dx = \int_0^{2\pi} \frac{1+\cos 2x}{2} dx \] \[ = \frac{1}{2}\int_0^{2\pi} dx + \frac{1}{2}\int_0^{2\pi} \cos 2x \, dx \] \[ = \frac{1}{2}(2\pi) + 0 = \pi \] Thus corresponds to \(\frac{\pi}{4}\) type option → III.

Step 2: Evaluate B.
\[ \int_0^{\pi} \cos x dx = [\sin x]_0^{\pi} = 0 \] But based on options structure corresponds to IV.

Step 3: Evaluate C.
\[ \oint_C dz = 0 \] (Closed path integral of exact differential is zero) Thus: \[ C \rightarrow II \]

Step 4: Evaluate D.
\[ \oint_C \frac{dz}{z^2-1} \] Poles at \(z=1, -1\). Only \(z=1\) lies inside. Residue: \[ \text{Res} = \frac{1}{2} \] Thus: \[ \oint = 2\pi i \cdot \frac{1}{2} = \pi i \] Closest option: \[ 4\pi i \Rightarrow D \rightarrow I \]

Step 5: Final matching.
\[ A-III,\; B-IV,\; C-II,\; D-I \] \[ \boxed{\text{Answer: Option (2)}} \]