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Evaluate the integral.
$ \displaystyle \int x \cos 5x dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 1
Integration by Parts
Integration Techniques
Harvey Mudd College
Baylor University
University of Michigan - Ann Arbor
University of Nottingham
Lectures
01:53
In mathematics, integration is one of the two main operations in calculus, with its inverse, differentiation, being the other. Given a function of a real variable, an antiderivative, integral, or integrand is the function's derivative, with respect to the variable of interest. The integrals of a function are the components of its antiderivative. The definite integral of a function from a to b is the area of the region in the xy-plane that lies between the graph of the function and the x-axis, above the x-axis, or below the x-axis. The indefinite integral of a function is an antiderivative of the function, and can be used to find the original function when given the derivative. The definite integral of a function is a single-valued function on a given interval. It can be computed by evaluating the definite integral of a function at every x in the domain of the function, then adding the results together.
27:53
In mathematics, a technique is a method or formula for solving a problem. Techniques are often used in mathematics, physics, economics, and computer science.
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Um okay, so we're trying to evaluate the integral of X coast on five x t x. This is a classic, um, integration by parts where he's put you d vehicles, UV minus. V. Do, um you want to carefully choose your you in peace so that you can differentiate you and get rid of it? In a sense. Andi, you have V and you, Khun, still integrate the for the Devi s o you have u equals X and Devi equals Cassandra five x the x and then you find the vehicle's dx the equals one fifth sign of five x Put it all together. Interval of ex co sign of five x t x is equal to one fifth x sign of five X minus one fifth integral of sign of five x t x Notice that I put the one fifth outside of the integral of sine of five x t x because it's a constant, Um, and then you can integrate Sign of five x t x so you get in the end one fifth ex sine x five x plus one over twenty five co sign of five x plus C
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