Correct Answer - Option 1 :
\(\frac{{-1}}{{5}}(-1+i)\)
Concept:
cos (nθ) + (i × sin (nθ)) = einθ
sin (nθ) + (i × cos (nθ)) = i × e-inθ
Multiplication of imaginary numbers
i × i = -1
-i × i = 1
\({1\over i}=-i\)
Integaration of standard functions
\(\int \sin ax=\frac{{-\cos ax}}{{a}}\)
\(\int \cos ax=\frac{{\sin ax}}{{a}}\)
Calculation:
Given:
\(I = \mathop \smallint \limits_0^{\pi /2} \frac{{\cos 3x + i\sin 3x}}{{\sin 2x + i\cos 2x}}dx\)
We know that
cos (nθ) + (i × sin (nθ)) = einθ
sin (nθ) + (i × cos (nθ)) = i × e-inθ
\(I = \mathop \smallint \limits_0^{\pi /2} \frac{{\cos 3x + i\sin 3x}}{{\sin 2x + i\cos 2x}}dx\)
\(I = \mathop \smallint \limits_0^{\pi /2} \frac{{e^{i3x}}}{{e^{-i2x}}}dx\)
\(I = \mathop \smallint \limits_0^{\pi /2} \frac{{e^{i5x}}}{{i}}dx\)
Multiply and divide by i
\(I =- \mathop \smallint \limits_0^{\pi /2} i× e^{i5x}dx\)
\(I = -i\mathop \smallint \limits_0^{\pi /2} e^{i5x}dx\)
Repacing einθ = cos (nθ) + (i × sin (nθ))
\(I =-i \mathop \smallint \limits_0^{\pi /2}\cos 5x+(i× \sin 5x) dx\)
\(I =-i [\frac{{\sin5x}}{{5}}-i\frac{{\cos5x}}{{5}}]_0^{\frac{{\pi}}{{2}}}\)
\(I=-\frac{{i}}{{5}}[\sin5x-i× cos5x]_o^{\frac{{\pi}}{{2}}}\)
\(I=-\frac{{i}}{{5}}[(1-0)(0-i)]\)
\(I=-\frac{{i}}{{5}}[1+i]\)
\(I=\frac{{1}}{{5}}[1-i]=-\frac{{1}}{{5}}[-1+i]\)
