Prove the integration formula (4.67).

Name: Problem 18, Chapter 4: Applied Linear Algebra (Undergraduate Texts in Mathematics)
Uploaded: 2020-12-20T19:41:46-08:00
Duration: 8 min 13 s
Channel: Ekaveera Kumar
Description: Prove the integration formula (4.67).

Key Concepts

Differentiation under the Integral Sign

This technique involves differentiating an integral that depends on a parameter with respect to that parameter. It allows one to bring differentiation inside an integral, thereby helping to establish connections between the integral and its derivative form, which can be key in proving general integration formulas.

Change of Variables (Substitution)

This method involves replacing the variable of integration with another variable to simplify the integral’s form. By appropriately choosing a substitution, one can transform a complicated integral into a standard form, making it easier to evaluate or compare to known formulas.

Fundamental Theorem of Calculus

This theorem provides the bridge between differentiation and integration by stating that if a function is continuous over an interval, then its definite integral can be calculated via its antiderivative. It is the backbone of proving many integration formulas, as it allows one to evaluate integrals by understanding their corresponding derivative processes.

Integration by Parts

Derived from the product rule for differentiation, integration by parts is a technique that converts the integral of a product of functions into simpler integrals. It is often instrumental in proving integration formulas by re-expressing integrals in forms that are easier to evaluate, thereby revealing underlying identities.

Transcript

00:01 Let's derive the integration formula 7 which is given by integration of du by u root u square minus a square is 1 by a secant inverse modulus of u by a plus some constant of integration.

00:15 And of course this formula is valid when a is not equal to 0.

00:19 I'll give you the best proof.

00:22 To find this integral, let's find its derivative first.

00:27 Let's see how derivative of secant inverse is defined.

00:30 We know that when y is equal to secant inverse x the principal range of y is basically belonging to 0 to pi this is the principal range of second inverse x that means the y will be either in first quadrant or in second quadrant now when you take secant on both sides i'll get second y is equal to secant of secant inverse x that implies secant y is nothing but x now now let's differentiate both sides.

01:10 What is derivative of second y? it is second y tan y into dyy is equal to d x.

01:18 So that means my dy by d x is nothing but 1 by secant y into tan y.

01:27 Now comes the very important point case 1.

01:31 When y belongs to 0 to pi by 2, i'm putting in open interval because tan y will not be defined when i close 0 or pi by 2.

01:41 So when y belongs to 0 to pi pi by 2 my tan y is positive but what is my tan y it is root of second square minus y but what is my second y my second y is x so it will be root of x square minus 1 so that means in case 1 my d by d x is nothing but 1 by x root x square minus 1 so this is only in the first case but whereas coming to the second case what is second case when y belongs to pi by 2 to pi this is in second quadrant my tan y should be negative so it's minus root of second square by minus 1 which is minus root of x square minus 1 so in this case my d y by d x will be 1 by x into minus root x square minus 1 or simply 1 by minus x root of x squared minus 1.

02:48 Now see the final conclusion.

02:52 I can combine the derivative in case 1 in case 2 something like this.

02:56 So, d by d x of second inverse x will be 1 by modulus of x root of x square minus 1.

03:08 So why modulus came into picture? i am clubbing the minus x and the x part as a modulus of x.

03:16 That means when x is greater than 1, i'll be getting this part.

03:23 When x is less than minus 1, i'll get this part...

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