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Evaluate the integral.

$ \displaystyle \int t \csc^2 t dt $

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$-t \cot t+\ln |z|+C=-t \cot t+\ln |\sin t|+C$

Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Missouri State University

Oregon State University

Harvey Mudd College

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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Find the indefinite integr…

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Evaluate the indefinite in…

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Find the general indefinit…

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Evaluate the integrals…

Evaluate each integral.

The problem is evaluated the integral t times per second t square dt. This problem. We will use the method of integration by parts, so the formula is integral: u from dx is equal to: u times, v minus the integral of u prime v d x. Now for our problem, we can let? U is equal to p and the prime is equal to second square. Then u, prim is equal to 1 and v is equal to negative cotananttnow, integral o f t times. Second t square dt is equal to u times so this is negative to tangent t minus integral of 1 times negative cotangent t, but this is equal to negative 2 tandant t minus this is past minus minus co. Tangent to this pass integral of co tangent to we can write cotangent t as cosine t over sine t t now for this integral. We can use? U substitution, that? U is equal to sine t, then du equal to cosine t, so this integral is equal to integral 1 over. U u! This is not. Of this is absolute value of sine t now integral of t cosecant square. This is equal to negative t. Co, tangent t plus in absolute value of sine t and pass con number c.

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