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Evaluate the integral.
$ \displaystyle \int \cos^{-1} x dx $
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Calculus 2 / BC
Chapter 7
Techniques of Integration
Section 1
Integration by Parts
Integration Techniques
Harvey Mudd College
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.
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Evaluate the integral.…
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Evaluate the following int…
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Evaluate the indicated int…
The problem is evaluated the integral integral cosine numerous x dx. For this problem we will use the method of integration by parts, so the formula is integral: u v, primedxis equal to? U times, v, minus integral? U, prime with the x now for this problem. We can let? U is equal to a sine us x and the last from is equal to 1. Then we have. U, prime, is equal to negative 1 over root of 1 minus x square and v is equal to x. Now use this formula. We have this integral is equal to cosin, inverse x times x, minus the integral negative 1 over root of 1 minus x square times x dx. This is equal to x, cosine, universe, x, plus integral x, over root of 1 minus x squared dx. For this part, we can use the method of u substitution and let? U is equal to 1 minus x square and d? U is equal to negative 2 x dx, so this integral is equal to negative, integral 1 half times u to negative 1 half du. This is equal to negative: u to 1 half. This is negative root of 1 minus x square. Now we know the integral of cosine inverse x dx. This is equal to x times, cosine numerus x and then, as the integral is 1, that minus the root of 1 minus x square, and it don't forget, as of constant number.
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