GATE CS 2009 Question
Duration: 4 min
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The video features an educational lecture by Yash Jain Sir from Knowledge Gate, focusing on solving a definite integral problem from the GATE CS 2009 exam. The core task involves evaluating the integral of (1 - tan x) / (1 + tan x) from 0 to pi/4. The instructor demonstrates a substitution method to simplify the trigonometric expression into a standard logarithmic integral form. The session emphasizes algebraic manipulation of trigonometric identities and the recognition of derivative patterns within integrands. This problem serves as a practical example of applying calculus techniques to competitive exam questions.
Chapters
0:00 – 2:00 00:00-02:00
The session begins with a title card displaying "CALCULUS" surrounded by mathematical formulas. The instructor introduces the problem: integral from 0 to pi/4 of (1 - tan x) / (1 + tan x) dx. The multiple-choice options A, B, C, and D are listed on the left side of the screen. He writes the integral on the whiteboard and begins the simplification process. He substitutes tan x with sin x / cos x. By multiplying the numerator and denominator by cos x, he transforms the integrand into (cos x - sin x) / (cos x + sin x). He identifies the numerator as the derivative of the denominator, setting up a substitution where u = cos x + sin x. This step is crucial for converting the complex trigonometric fraction into a manageable algebraic form.
2:00 – 4:16 02:00-04:16
The instructor proceeds with the substitution method. He differentiates u to find du = (cos x - sin x) dx, confirming the numerator matches du. He then changes the limits of integration: when x=0, u=1; when x=pi/4, u=sqrt(2). The integral becomes integral from 1 to sqrt(2) of (1/u) du. He evaluates this to [ln|u|] from 1 to sqrt(2), resulting in ln(sqrt(2)) - ln(1). Finally, he simplifies ln(sqrt(2)) to (1/2)ln(2) and circles option D as the correct answer. The visual progression on the board clearly shows the transformation from the original problem to the final logarithmic solution. He explicitly writes out the evaluation steps, ensuring clarity for the students.
The lecture effectively demonstrates a standard technique for solving trigonometric integrals involving tangent functions. By converting tangent to sine and cosine, the problem simplifies into a form suitable for u-substitution. The key insight is recognizing the derivative relationship between the numerator and denominator, which allows for a direct integration into a natural logarithm function. This method is a classic approach for GATE-level calculus problems, showcasing how algebraic simplification can reveal underlying calculus structures. The instructor's methodical approach ensures that students understand not just the answer, but the logical flow of the solution.