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Evaluate the integral.

$ \displaystyle \int (\arcsin x)^2 dx $

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$x(\arcsin x)^{2}+2(\arcsin x) \sqrt{1-x^{2}}-2 x+C$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Harvey Mudd College

Baylor University

University of Nottingham

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.

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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 indefini…

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Calculate the integrals.

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Evaluate the indefinite in…

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Calculate.$$\int \frac…

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The problem is find the integral of a tenant sine x, squared dx, for this problem. We can use the method of integration by parts. The formula is integral of: u times, vois equal to? U times, v minus the integral of u prim times v x. For our problem, we can not? U is equal to k sine x square and prime is but 1, then you primis equal to 2 times tanantxyne x times 1 over root of 1 minus x square and v is equal to x. Now use this formula, this integral is equal to u times v, so this is x times oxyde x, squared minus integral of prim times, so this is to oxide x times 1 over. This is x over root of 1 minus x, squared dx. Now for this function for this integral, we can also use integration by parts that, u is equal to 2 times oxine x and prime, is equal to x over root of 1 minus x square, then you promis equal to 2 over root of 1 minus x square And v is equal to integral of prime, and this is equal to integral of x. Over root of 1 minus x square texts. 1. We kind o use! U substititution, that? U is equal to 1 minus x square, then du is equal to negative 2 x dx. So this is equal to integral negative 1 half times u over nuter. U, this is equal to negative rutifuand. U is equal to 1 minus x squared. This is equal to negative root. If 1 minus x, squarenow this integral this is equal to x, have oxide x. Squared minus this integral is so they use integration by powers. U times v. This is 2 oxy x times negative root of 1 minus x, squared minus the integral of primis way. So this is negative. 2, just negative 2 x. So this is equal to x, xi, x, square plus 2 times the roots of 1 minus. I squared x, inex- and this is minus minus minus of this and minus 2 x and and pass to constant the number. This is the answer for this integral.

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