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Verify Formula 31 (a) by differentiation and (b) by substituting $ u = a \sin \theta $.
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 6
Integration Using Tables and Computer Algebra Systems
Integration Techniques
Michael F.
July 28, 2021
Why did you sub in a sin (theta) for u at the start?
Campbell University
University of Nottingham
Idaho State University
Boston College
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.
01:02
Verify Formula 31 (a) by d…
01:48
Verify the identity.$$…
01:27
Okay, This question wants us to verify this integration formulas, so it gives us a hint to try and suggest the substitution u equals a sign data. So that would mean that do you is a co sign theta d thing. So now we can just plug in and get our integral of a sign. Data squared times the square root of a squared minus a sign data squared times a co signed data d theta. So now that everything is in terms of data, let's see what we can simplify. So we have and a squared in front and in athe end So we get in a cube sine squared data co signed data times the square root of a squared times one minus sine squared data d theta. So then we get a to the fourth time's the integral of sine squared data co signed data times the square root of co sign squared data d theta because we used the Pythagorean identity. And then finally, we're left with eight to the fourth times the integral of sine squared fada co sign squared fada d theta. So now we have a couple options as to what we can do to evaluate this. So remember that sign of tooth Ada equals to sign Fada co sign theta So sign of to fate over to equals Science data co sign data, but we have signs squared co sign squared data. So we need to square both sides. So that says sine squared of two fada over four equals sine squared data co sign squared data. So now let's plug that in. So we're left with eight of the fourth over four times the integral of sine squared of tooth ada d theta. Then we can use a u substitution. But in this case were to use to you. So we have to use V. So V is equal to tooth Ada. So Devi is equal to two d theta or Devi over too equals d theta. So then we have a to the fourth over eight times the integral of sine squared the devi and this is something we know the integral oath. So we get a to the fourth over four times V over too minus sign of two V over four plus c Just checking that in the table real quick. Yeah, and now we just have to back substitute and we're good. So eight of the fourth over four times. Well, what is vey vey is tooth Ada. So we get to fate over to we needed our chain roll factor. This should be an eight. So then this is still eight of the fourth over eight times Tooth Ada over to minus 1/4 sign of fourth Ada plus c. And then we have to back substitute for theta. So data is equal to sine inverse of you are a So then we get the following minus 1/4 sign of four. Sign in verse of you over a and using right triangle trigonometry. We can simplify this expression. So what it turns into is a to the fourth over a sign in verse of you over a plus You, over eight times to you squared minus a squared square root of a squared minus. You squared plus c. So this is our final formula
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