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Evaluate the integral.
$ \displaystyle \int x \tan^2 x dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 1
Integration by Parts
Integration Techniques
Oregon State University
Baylor University
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.…
The problem is evaluated the integral x times standard x, squared dx, for this problem. First, we can write that this integral is equal to integral x m. O n in the x squared is equal to second x squared minus 1 point. This is equal to integral of x times second x, squared dx minus the integral of x. Now we compute this integral to the integral of x times second x, squared dx, for this integral, we will use method of integration by powers. The formula is integral of u times. Prime dx is equal to: u times, v minus integral of? U prime, for this problem we can write. U is equal to x and the prime is equal to second x square. Then prime is equal to 1 and v is equal to tan, and now this integral is equal to u times b. So this is x times tan ant x, minus the integral you prim times v. So this is tangent x, dx minus the tenor x d x is 1 half x square and the integral of tenant x x. This is the integral of sine x over cosine x. We can use? U substitution that? U is equal to cosine, x and d? U is equal to negative sine x x, so this is equal to the integral of negative. 1 number? U d! U s! This is negative n have absolute value of u so this is absolute value of cosine x. So this is equal to x, tangent, x, minus minus negative point. So this is as an absolute value of cosine x, minus 1, half x square and plus constant number c.
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