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Evaluate the integral.

$ \displaystyle \int e^{-\theta} \cos 2 \theta d \theta $

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$\int e^{-\theta} \cos 2 \theta d \theta=\frac{1}{5} e^{-\theta}(2 \sin 2 \theta-\cos 2 \theta)+C$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Campbell University

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.

05:28

11:55

Evaluate the integral.…

07:50

02:01

Evaluate the integrals.

03:59

00:34

02:30

04:01

Use integration by parts t…

01:41

So the problem is evaluated the integral into negative theta times cosine 2 theta d theta. For this problem we will use the method of integration by parts. The formula is integral of: u times. Prime d x is equal to u times v minus the integral of our prime. For this problem we can let? U is equal to 2 negative theta and the atom a prime is equal to cosine 2 theta. Then your problem is equal to negative e to negative theta and the way is equal to while half times sine 2 times now. This integral is equal to u times v, so this is 1 half to negative theta sine 2 theta minus integral of u prim times. V, so this is plus 1 half into row of e to negative theta sine 2 theta d theta now located this integral. So this is similar with this 1, but we can also use integral integration by parts that, u is equal to 2 negative theta and promise to sine 2 times the then you promise to negative 2 negative, theta and v is equal to negative 1 half cosine 2 Times now this is equal to 1 half to negative theta sine 2 times theta plus this is 1 half times u times v. This is e to negative theta times negative 1 half cosine 2 times theta minus the integral of your prime times. This is integral of 1 half times e to negative theta cosine 2 theta d. This is equal to 1, half to negative theta sine 2 times theta and then minus 1 force into negative theta, cosine, 2 theta and minus 1 force integral of into negative theta cosine. 2 theta tap now look at this integral. This is the same as this integral. So we can move this term to the right left hand side and we have 1 plus 14 times integral of into negative theta cosine 2 theta d theta. This is equal to 1. Half to tie theta sine 2 theta minus 1 for to negative theta cosine 2 times, since this number is 5 over 4. So we multiply 4 over 5 for both sides and we have integral of e to negative theta cosine 2 theta d theta. We multiply 54 over 5, so this is 2 over 5 into negative theta sine 2 theta minus 1 over 5 into negative theta cosine, 2 theta and plus constant num.

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