GATE Question on Integration by Parts Method
Duration: 7 min
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This educational video features a calculus lecture by Yash Jain Sir, focusing on solving a definite integral problem from the GATE examination. The core task is to evaluate the integral of x squared times cosine x from 0 to pi. The instructor systematically applies the integration by parts technique, utilizing the LIATE rule to determine the choice of u and dv. The lesson covers setting up the integral, evaluating boundary terms, and handling a secondary integral that arises from the first application of the method.
Chapters
0:00 – 2:00 00:00-02:00
The video begins with an intro screen displaying CALCULUS surrounded by various mathematical formulas. The instructor introduces the problem integral from 0 to pi of x squared cos x dx written on the whiteboard. He lists essential trigonometric values needed for evaluation: cos 0 equals 1, cos pi equals negative 1, sin 0 equals 0, and sin pi equals 0. He then introduces the LIATE rule Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential to guide the selection of parts for integration by parts. He identifies x squared as the Algebraic part u and cos x as the Trigonometric part dv. He writes the standard integration by parts formula integral uv equals u integral v minus integral u prime integral v.
2:00 – 5:00 02:00-05:00
The instructor proceeds to apply the formula. He calculates the integral of cos x as sin x. Substituting these into the formula, he writes bracket x squared sin x bracket from 0 to pi minus integral from 0 to pi of 2x sin x dx. He evaluates the first term bracket x squared sin x bracket from 0 to pi by substituting the limits, noting that sin pi and sin 0 are both zero, causing this entire term to vanish. This leaves the remaining integral negative integral from 0 to pi of 2x sin x dx. He factors out the constant 2 to get negative 2 integral from 0 to pi of x sin x dx. He prepares to apply integration by parts again to this new integral, identifying x as u and sin x as dv. He writes the integral of sin x as negative cos x.
5:00 – 6:51 05:00-06:51
The instructor completes the second integration by parts. He writes the expression negative 2 bracket x negative cos x minus integral 1 negative cos x bracket. He simplifies the inner integral to integral cos x, which evaluates to sin x. The full expression becomes negative 2 bracket negative x cos x plus sin x bracket from 0 to pi. He evaluates the limits at pi and 0. At the upper limit pi, the term becomes negative 2 bracket negative pi cos pi plus sin pi bracket, which simplifies to negative 2 bracket negative pi negative 1 plus 0 equals negative 2 pi. At the lower limit 0, the term is zero. The final result is negative 2 pi. The video concludes with a THANKS FOR WATCHING screen.
The lecture effectively demonstrates a multi-step integration problem. By breaking down the process into two distinct applications of integration by parts, the instructor shows how to reduce the power of the polynomial term x squared until it disappears. The use of the LIATE rule ensures the correct choice of functions, preventing unnecessary complexity. The final evaluation relies heavily on the specific values of trigonometric functions at the boundaries 0 and pi.