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Evaluate the integral.

$ \displaystyle \int t^2 \sin \beta t dt $

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03:58

Wen Zheng

05:38

Carson Merrill

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Baylor University

University of Michigan - Ann Arbor

University of Nottingham

Idaho State University

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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Evaluate the integral.…

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Evaluate the definite inte…

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Find the indefinite integr…

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Evaluate.$$\int \f…

to evaluate this integral. We first apply substitution and here we let X equal to beta times T. And so he is just X over beta and D T would be one over beta times dx. And so from here we have the integral of X over beta squared times Sine of X Times one over beta DX. And then simplifying we have one over beta to the third power times the integral of x squared times sine of X the X. Now for this integral we will apply integration by parts. In here we let U equal to X squared and DV equal to sine of X dx. And so from here we have do you Which is two XDX and the equal to negative cosine of X. And so we have one over data to the third power This times U times V that's negative X squared co sign of X minus the integral of VD you that's negative Cosine of X Times two X DX. And simplifying we have one over Beta to the 3rd power times negative X squared cosine of X plus two integral of X. Cosine of X dx. And then for this integral will apply integration by parts. Again in here we let U equal to X and DV equal to cosine of X dx. And so we have do you echoes dx and V which is just sign of X. So from here we have one over beta to the third power times negative X squared times cosine of X Plus two times u times v That's X times sine of X minus the integral of E d U. That's sine of X dx. So simplifying, we have won over beta to the third power times negative X squared times Cosine of X plus two X sine of X minus two times negative co sign effects. And then plus C. So we find further that's trust one over beta to the third power times negative X squared times school side effects Plus two x sine of x plus co sign of X. And then plus C. And since access beta times T we have one over beta to the third power times negative of beta times T squared times co sign of beta times T plus two times beta times T sine of beta time steve let's go sign of beta times T. And then plus C. That means we have negative one over beta times we have t squared co sign of beta times T Plus We have two over beta squared times T times sine of beta times T Plus one over beta to the third power times co sign of beta, T blast. See and so this is our integral one over paid up to the 3rd power this times negative X squared co sign of X plus two x. sine of x plus to co sign of X and then plus C. And because excess beta times T we have won over beta to the third power times negative of beta times T squared co sign of beta times T last two times. Beta times T times sine of beta times T plus to co sign of beta times T plus C and then distributing one over beta through the third power we have negative one over beta squared -1 over beta times T squared co sign of wait A times T plus. We have to over beta squared times T times sine of beta times T Plus two over beta to the third power times, co sign of beta times T and then plus C. And so this is the indefinite integral.

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