GATE CSE 2011 | Question: 31

Question

Given $i = \sqrt{-1}$, what will be the evaluation of the definite integral $\displaystyle \int \limits_0^{\pi/2} \dfrac{\cos x +i \sin x} {\cos x - i \sin x} dx$ ?

• A. $0$
• B. $2$
• C. $-i$
• D. $i$

Comments

f(X)

---

• $e^{i\theta} = \cos\theta + i \sin\theta$
• $e^{-i\theta} = \cos\theta - i \sin\theta$

Reference:

• https://math.stackexchange.com/questions/2735303/does-eulers-formula-give-e-ix-cosx-i-sinx 

---

cos 180 = cos(90 + 90) = - sin 90 = -1

School maths

90, 270 => Change(sin <=> cos, tan <=> cot, sec <=> cosec)

Sign, 

1st Quadrant(90-x,) = all positive

2nd(90+x, 180-x) = sin, cosec

3rd (180+x, 270-x)= tan, cot

4th (270+x, 360-x) = cos, sec

---

.

Answer 1

$$
\begin{aligned}
\int_{0}^{\pi/2} \frac{e^{ix}}{e^{-ix}} \, dx
&= \int_{0}^{\pi/2} e^{2ix} \, dx \\

&= \left[ \frac{e^{2ix}}{2i} \right]_{0}^{\pi/2} \\

&= \frac{e^{i\pi} - 1}{2i} \\

e^{i\pi} &= \cos \pi + i \sin \pi \\

&= -1 + i(0) \\

\therefore \quad
\frac{(-1)-1}{2i}
&= \frac{-2}{2i} \\

&= \frac{-1}{i} \\

&= i
\end{aligned}
$$

Answer 2

file:///C:/Users/ram/Desktop/20180728_151152.jpg

Comments

OK