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Use integration by parts to prove the reduction formula.

$ \displaystyle \int \tan^n x dx = \frac{\tan^{n -1} x}{n - 1} - \displaystyle \int \tan^{n -2} x dx (n \neq 1) $

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Calculus 2 / BC

Chapter 7

Techniques of Integration

Section 1

Integration by Parts

Integration Techniques

Oregon State University

Harvey Mudd College

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.

03:46

Use integration by parts t…

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Let $n>2 .$ Establish t…

To prove the problem is prove the reduction formula integral of time it x to x power, that's equal to 10 to 1 power over m minus 1 minus the integral of an ant x, 2 minus 2 power x. For this problem, i don't use integration by parts. Instead, i use us substititution, let's see how to do it here, i need to trick identity. Second x, squared minus 1 is equal to tan ant x square. Now the integral of pendant is power. X, dx is equal to integral of an ant x, 2 minus 2 times an ant x square right second x, squared minus 1. This is equal to integral of tandant x. 2, minus to this power have second x, squared dx, minus integral of tandant x, 2 and minus 2 x, and for this integral we can let? U is equal to an ant. Then du is equal to second x squared d. Now this integral is equal to integral of u to n minus 2 power? U so! This is equal to 1 over n 1! U 2 and minus 1 point! This is addressed. The first to this is 10 x, 2 and minus 1 over n minus 1 point, and the this term is just minus the integral of panax 2 minus those power. This is the res.

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