GATE CS 2011 – 2 Marks Question

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The video presents a solution to a definite integral problem from GATE CS 2011 involving complex numbers. The integral is $\int_{0}^{\pi/2} rac{\cos x + i \sin x}{\cos x - i \sin x} dx$. The instructor demonstrates two distinct methods to solve it. Method 1 utilizes Euler's formula to convert trigonometric functions into exponential form, simplifying the integrand to $e^{2ix}$. Method 2 involves rationalizing the denominator using the conjugate, applying double-angle identities, and integrating the resulting real and imaginary parts separately. Both methods yield the final answer $i$.

Chapters

1. 0:00 – 2:00 00:00-02:00

   The instructor introduces the problem statement: "Given $i = \sqrt{-1}$, what will be the evaluation of the definite integral $\int_{0}^{\pi/2} rac{\cos x + i \sin x}{\cos x - i \sin x} dx$?" He identifies it as a GATE CS 2011 question worth 2 marks. He begins Method 1 by writing down Euler's formulas on the right side of the board: $e^{i heta} = \cos heta + i \sin heta$ and $e^{-i heta} = \cos heta - i \sin heta$. He then starts substituting these into the integral expression, pointing to the numerator and denominator to show the correspondence with the exponential forms. He writes the integral with the exponential substitution $\int_{0}^{\pi/2} rac{e^{ix}}{e^{-ix}} dx$. The text "Method 1" is visible in the top left corner.

2. 2:00 – 5:00 02:00-05:00

   The instructor simplifies the integrand using the exponential forms, writing $rac{e^{ix}}{e^{-ix}} = e^{2ix}$. He sets up the integral $\int_{0}^{\pi/2} e^{2ix} dx$ and finds the antiderivative $\left[ rac{e^{2ix}}{2i} ight]_0^{\pi/2}$. He evaluates the limits, substituting $x = \pi/2$ to get $e^{i\pi}$ and $x=0$ to get $e^0$. He writes $e^{i\pi} = \cos \pi + i \sin \pi = -1$. The calculation proceeds to $rac{-1 - 1}{2i} = rac{-2}{2i} = rac{-1}{i}$. He rationalizes this to get the final result $i$. He explicitly writes out the steps $rac{-1}{i} imes rac{i}{i} = rac{-i}{i^2} = rac{-i}{-1} = i$. The board shows the full integration process from start to finish.

3. 5:00 – 10:00 05:00-10:00

   The instructor switches to Method 2. He writes the original integral and multiplies the numerator and denominator by the conjugate $(\cos x + i \sin x)$. He expands the numerator to $\cos^2 x + i^2 \sin^2 x + 2 \cos x \cdot i \sin x$. He simplifies the denominator to $\cos^2 x - i^2 \sin^2 x = \cos^2 x + \sin^2 x = 1$. He uses the identities $\cos^2 x - \sin^2 x = \cos 2x$ and $2 \sin x \cos x = \sin 2x$ to rewrite the integrand as $\cos 2x + i \sin 2x$. He then integrates this expression: $\int_{0}^{\pi/2} \cos 2x dx + i \int_{0}^{\pi/2} \sin 2x dx$. He evaluates the antiderivatives $\left[ rac{\sin 2x}{2} ight]_0^{\pi/2} - i \left[ rac{\cos 2x}{2} ight]_0^{\pi/2}$. The text "Method 2" is visible in the top left corner.

4. 10:00 – 10:32 10:00-10:32

   The instructor completes the evaluation for Method 2. He substitutes the limits into the trigonometric expressions: $\sin \pi - \sin 0$ and $\cos \pi - \cos 0$. He calculates the values: $\sin \pi = 0$, $\sin 0 = 0$, $\cos \pi = -1$, $\cos 0 = 1$. The expression simplifies to $0 - i rac{-1 - 1}{2} = -i rac{-2}{2} = i$. This confirms the result obtained in Method 1. The video concludes with a "THANKS FOR WATCHING" screen.

The video provides a comprehensive tutorial on evaluating a complex definite integral using two distinct mathematical approaches. Method 1 leverages the power of Euler's formula to transform trigonometric functions into exponential form, significantly simplifying the integration process. Method 2 relies on algebraic manipulation, specifically rationalizing the denominator and applying double-angle trigonometric identities. Both methods converge on the same result, $i$, reinforcing the consistency of complex number operations. This dual-method approach is valuable for students preparing for competitive exams like GATE, as it demonstrates flexibility in problem-solving strategies and deepens understanding of complex analysis concepts.