The correct answer is option 4.Calculation:\(\displaystyle\int^{\pi/2}_0{\dfrac{\cos x+i\sin x}{\cos x-i \sin x}}dx\)According to Eulers formula, \(e^{i?}? = cos ? + i sin ? \\ e^{-i?}? = cos ? - i sin ? \)Therefore\(e^{ix} = cos x+ isinx \\ e^{-i?}? = cos ? - i sin ? \\ I=\int_{0}^?{ e^{ix} \over e^{-ix} } dx \\ I =\int_{0}^{? \over 2}{ e^{i2x} } dx\)\( \\ I = [{ e^{i2x} \over 2i}]_{0}^{? \over 2} \\ I= {1 \over 2i}[cos2x+isin2x ]_{0}^{? \over 2} \\ I= {1 \over 2i} \times (-2) \\I={ -1 \over i } \\I=i \space \space \space \space \space \space \space \space (? {1 \over i} =-i) \)Hence the correct answer is i.