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Evaluate the integral.

$ \displaystyle \int \sin 8x \cos 5x dx $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 2

Trigonometric Integrals

Integration Techniques

Baylor University

University of Michigan - Ann Arbor

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 integral from…

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This problem is from Chapter seven section to problem number forty one in the book Calculus Early Transcendental Zane Condition by James Door and we have on indefinite inaugural of Sign of a Dicks Times Co signed by VIX and R inte Grant is of this form signed eight times co sign be in our keys. Our value may is in X in Our value of B is five. Six so we can rewrite the integral using this formula up here. So doing so, we have one half in a rule sign of a minus B. So we have eight x minus five bricks plus sign of a plus thie, which is in X plus five mics. So let's go ahead and simplify this the sign of three X and sign of thirteen x. So, in my help here to use use up if you'd like, you can go ahead and take you too. Be free X in here. We could take you to be thirteen x So evaluating these inner rules, we should obtain negative co signed three x over three minus co signed thirteen x over thirteen and plus our constant of integration and the next thing we could do is just simplify this a little bit. Clean this up. We have a negative co signed three X over six, minus code sign thirteen x over twenty six, plus he and that's our answer.

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