Evaluate the integrals in Exercises 67-74 in terms of a. inverse hyperbolic functions. b. natura

Evaluate the integrals in Exercises 67-74 in terms of a....

Key Concepts

Partial Fraction Decomposition

This technique involves breaking down a complex rational function into a sum of simpler fractions that are easier to integrate. It is a fundamental tool in integration, especially when the denominator factors into linear or quadratic terms, allowing one to apply standard antiderivative formulas.

Inverse Hyperbolic Functions

Inverse hyperbolic functions, such as arccoth or artanh, arise naturally as antiderivatives of functions with quadratic denominators. They provide a convenient form for expressing antiderivatives when the integrand matches the derivative of an inverse hyperbolic function, and they carry essential domain restrictions that must be observed.

Natural Logarithms in Integration

The natural logarithm function often appears as an antiderivative when integrating expressions of the form 1/u. Many antiderivatives, especially those derived after partial fraction decomposition, can be rewritten in logarithmic form. This linkage is especially important when converting between inverse hyperbolic functions and logarithmic expressions.

Definite Integration and Domain Considerations

Evaluating a definite integral requires applying the antiderivative at the specified bounds and subtracting. When working with inverse hyperbolic functions and logarithms, it is crucial to consider the domains of these functions, as the validity of the antiderivative expression may depend on the range of the integration variable.

Transcript

00:02 In this problem, we have the integral of 5 -4s, from 5 -4s to 2, of the x over 1 minus x squared.

00:14 Well, just by definition, this integral is the inverse hyperbolic cotangent of x.

00:20 Well, not by definition.

00:22 You can derive it, but it's derived in the chapter.

00:25 So you can just use this identity here.

00:30 So all we need to do is, um, is.

00:36 Plug in the bounds.

00:38 So we get the inverse hyperbolic cotangent of two minus the inverse hyperbolic cotangent of four -fifths.

00:44 Now, because these two values are greater than one, we can write that in terms of natural logs and still have it be a real number.

00:55 So to do that, again, this term here winds up being one half times the natural log of three over one.

01:03 And this term winds up being um um um oops i think i yeah i have a i should have a minus sign in here i'm not sure why it wrote a plus so then this term winds up being one -half times the natural log of nine -fourths over one -fourth so we get um one -half the natural log of three minus one -half the natural log of nine this is um natural log of three so we get one -half the natural log of three minus the natural log of 3, which is minus 1 half the natural log of 3.

01:41 Or if we want to pull the minus sign in, we get 1 half times a natural log of 1 3rd.

01:51 This one, we integrate from 0 to 1 half, dx over 1 minus x squared.

01:58 And again, we can use the identity that this integral here is the inverse hyperbolic tangent of x.

02:06 I don't know why i wrote you of x.

02:11 And so we just need to evaluate it at the bounds.

02:15 The inverse hyperbolic tangent of zero is zero.

02:20 So this just becomes the inverse hyperbolic tangent of one -half.

02:25 Now, since one -half and zero are both less than one, so these bounds are in the range where we can convert this into a natural log, a function with natural logs, and again not have any worry about.

02:44 Imaginary numbers, we get that that is the one half times the natural log of three halves all over one half.

02:52 And that's just three.

02:54 So we get one half times the natural log of three.

02:58 Or we could write it as a natural log of three if we wanted to.

03:05 And this last one is a little more complicated, but not really.

03:10 We just, again, we're going to wind up using identities that are, you know, drive for us in the chapter...

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