Finding an Indefinite Integral In Exercises 15-46 , find the indefinite integral. ∫(tan(2 / t))/

Finding an Indefinite Integral In Exercises 15-46 , find the...

Key Concepts

Indefinite Integral

An indefinite integral is the collection of all antiderivatives of a function. It represents the reverse process of differentiation and includes an arbitrary constant since differentiation of a constant results in zero.

Integration by Substitution

Integration by substitution is a technique used to simplify an integral by changing the variable of integration. When an integrand contains a function and its derivative, a substitution allows the integral to be rewritten in a simpler form that is easier to evaluate.

Chain Rule

The chain rule is a fundamental concept in differentiation which provides a method for finding the derivative of composite functions. In integration, especially when applying substitution, understanding the chain rule helps identify how the differential of the new variable relates to the original variable.

Transcript

00:02 Okay, so in order to solve this integral, since it is not one of our commonly known ones, we're going to have to do some substitutions.

00:12 So my first attempt, when i'm trying to figure out what i want to use to substitute, i will set u equal to whatever is on the inside.

00:23 So 2 over t is the closest to the inside.

00:28 So that's what i'm going to pick.

00:29 I could rewrite this as u equals 2 times t to the negative 1.

00:38 And that's going to make it easier for me to take my derivative.

00:40 So when i take my derivative of you in terms of t, i'm going to bring this negative 1 down, and that leaves them with t to the negative 2.

00:53 So i'm going to have negative 2, t to the negative 2, which is really just negative 2.

01:05 Over t squared.

01:10 It's an equal sign.

01:11 So now i need to solve this for, oops, i forgot my dt up there.

01:19 There we go.

01:20 Now i need to solve this for dt so that i can substitute this in as well.

01:27 So i'm going to cross multiply, and i'm going to end up with dt times negative 2 equals du times t squared so dt is equal to negative 1 1⁄2 du times t squared now i'm ready to substitute back into my original so i'm going to have the integral of the tangent of you over t squared times now i'm substituting for dt now as well, one half, sorry, negative one half, d u times t squared.

02:30 And as we had hoped, since we're switching to use, all the t's will disappear.

02:36 Those will cross out.

02:38 And my negative one -half, since it's a constant, can just go sit in the front.

02:46 So now i'm taking the integral, i'm type negative one -half times the integral, of the tangent of u d u and there it is so tangent is not an obvious um an obvious integral so i'm going to do a substitute first i'm going to do a transformation and then i'm going to do another substitution so my negative one half isn't going away yet but i can change tangent into sign and which for a lot of us we're more familiar with those anyway.

03:28 So i have the sign of you over the cosine of you.

03:38 And if there's nothing on the inside, like there isn't here, then my next step for finding what i want to be to substitute is i will look for what's on the bottom of my fraction.

03:49 So let's try that.

03:50 I say v is equal to cosine of you.

03:57 And when we take the derivative of v in terms of u, we're going to have, well, the derivative of the cosine is the negative sign of you...

Please give Ace some feedback