Evaluate the integrals. ∫2 / √(3)^2 t sec^-1 t d t |...

Name: Problem 19, Chapter 8: Thomas Calculus
Uploaded: 2020-09-06T18:59:21-08:00
Duration: 8 min
Channel: Kyler Gray
Description: Evaluate the integrals. ∫2 / √(3)^2 t sec^-1 t d t

Key Concepts

Integration by Parts

Integration by parts is a technique derived from the product rule of differentiation that is especially useful when integrating the product of two functions. It involves selecting one part of the integrand to differentiate and another to integrate, thereby transforming the original integral into one that is hopefully easier to evaluate.

Inverse Trigonometric Functions

Inverse trigonometric functions, such as the arcsecant function, are the inverses of the standard trigonometric functions. Their properties and derivatives are often used in integration, and understanding them is crucial when they appear in the integrand. The derivative of the arcsecant function, for example, provides insight into how to handle integrals involving this function.

Definite Integration

Definite integration involves calculating the accumulation of quantities between given bounds. When applying techniques such as integration by parts or substitution, it is important to correctly evaluate the antiderivative at the upper and lower limits to obtain the final value of the integral.

Transcript

00:01 For problem 19, we need to calculate this definite integral using integration by parts.

00:07 So we're going to set u equal to the arc secant of t, and that will make du equal to 1 over t times the square root of t squared minus 1.

00:22 And depending on what class hearing, you may need to memorize this formula, or you may just be able to use it as a reference sheet, but this is something that comes up relatively often.

00:31 And then we have dv, which will equal t d t, which makes v equal to 1 .5t squared.

00:42 So we can now go ahead and piece these together.

00:45 So we'll leave off the limits of integration for now and just find the integral and then put them back in.

00:51 So that will give us 1 .5t2 times the arc secant of t.

00:59 So there we go.

01:00 That's the first part of our integration by parts.

01:03 And then we need to subtract the integral of v -d -du.

01:08 So we'll pull out that one -half, but that will give us t -squared over t times the square root of t -squared minus 1 d -t.

01:19 So fortunately, these t -s will cancel, so that will give us one -half t -squared, arc -sacet of t -minus one -half times the integral of t -over the square root of t -squared minus 1.

01:35 D t and we can make you actually we're going to use something other than you since we've already used you uh we're going to use w will equal t squared minus one which means d w will equal to t d t and if we solve for t d t as that's what we actually have so t d t will equal one half d w uh so now we can conveniently actually do the substitution.

02:03 So we'll get one -half t squared times the arc secant of t, and we'll plot this one -half, so minus one -fourth, right? multiply those one -halfs together times the integral of one with the square root of w -d -w.

02:21 And we're going to actually rewrite this instead of one over the square root of dw as just w to the negative one -half power.

02:30 So that way it's easier to integrate.

02:31 So lastly, we'll do one -half t squared times the arc secant of t -minus.

02:39 And so now we have one -fourth times add one to that power.

02:44 So we get w plus one to this, and that becomes just positive one -half.

02:48 Then we need to divide by its stream power, so multiply by two plus c.

02:53 Actually, we won't use that plus c because we'll add back to limits of integration here.

02:57 So there we go.

02:58 So lastly, let's just distribute this real quick.

03:00 One -half t squared, arc secant of t and then minus one half w to the one half power or we can use square roots now and we'll also back substitute here so we'll just skip a step but a square root of t squared minus one there we go and now we can add our limits of integration which are two and two over root three so two and two over square root of three so there we go we're getting closer we're now we just need to actually plug in these limits and see where they lead us.

03:38 So let's start doing just that.

03:41 So let's plug in two first.

03:43 So we'll get one half times two squared, which is four.

03:46 So four over two is just two times the arc secant, arc secant of two minus one.

03:56 So two -half times t squared minus one.

03:57 So t squared minus one will be two squared.

04:00 So four minus one is three.

04:02 So root three over two.

04:05 Two.

04:05 There we go.

04:06 So that's our first segment...