Integrate using the method of trigonometric substitution. Express the final answer in terms of t

step 1 So let us have this definite integral right here. 1 minus x to the power. Okay, so we have this

Integrate using the method of trigonometric substitution....

Key Concepts

Trigonometric Substitution

Trigonometric substitution is a technique used to convert integrals containing square roots of quadratic expressions into integrals involving trigonometric functions. This method leverages identities such as sin²? + cos²? = 1 to simplify the integrand, making it easier to evaluate the integral.

Change of Variables in Definite Integration

Changing the variable in a definite integral involves substituting a new variable for the original one and also adjusting the limits of integration accordingly. This approach not only simplifies the integrand but also ensures that the integration is performed over the correct interval in the new variable.

Integration of Powers of Trigonometric Functions

Integrals that contain powers of trigonometric functions often require the application of trigonometric identities or reduction formulas to simplify the expression. Mastery of these techniques is essential for efficiently evaluating the resulting integrals after a trigonometric substitution is applied.

Symmetry in Integrals

Recognizing symmetry in the integrand can simplify the evaluation of definite integrals, especially when the function is even or odd over symmetric limits. By leveraging the even or odd nature of a function, one can often reduce the computation to a simpler form or a smaller interval.

Transcript

00:01 So let us have this definite integral right here.

00:08 1 minus x to the power.

00:17 Okay, so we have this one.

00:19 You can go through what we've been doing in previous tutorials, like two or three tutorials before by drawing the things and all that.

00:28 It's not a problem.

00:29 But straight away, you can see that here.

00:34 We're going to let x be sign u okay sign you then d x is going to be cosine u okay after you then d x is going to be cosine u okay d u right okay so uh see we can we can change the limits of integration okay after you make a substitution you can change the limits of integration or you can just continue to work like that and then change it back but for this purposes i'm gonna change it back so this is a definite integral so at x equals negative 1 then what is you gonna be if x is negative 1 then you gonna have negative 1 equals you right so you is going to be inverse sign of negative one so what is inverse sign of negative one negative pi over two right so negative pi over two and then at x equals positive one then you have positive one equals sign you so you is going to be inverse sign right inverse sign of one so what is inverse sign of 1? that is going to be pi over 2, right? so this is pi over 2 right here.

02:30 Okay, so we're going to replace this limits of integration here by that of a new variable.

02:38 Okay, so this one is going to be pi over 2 and then negative pi over 2 right here.

02:46 And then wherever i see x, i'm going to put sine u.

02:50 So this is going to be signs squared u, right? because i saw x, okay? and then here's my 3 over 2.

02:58 And then my d x is this guy right here.

03:01 Okay, so this is cosine u, d u, okay? good.

03:07 So i put this one in this situation.

03:11 Then i can simply solve the problem.

03:15 I'm going to make another substitution, okay? so if i make another substitution here, you know, 1 minus sine u is actually cosine u.

03:35 Okay, so this integration, i'm going to make that substitution.

03:40 This integration is going to be cosine squared u to the power 3 over 2, right, and then cosine u -d -u because 1 minus sine square u.

03:56 Is actually cosine square u right it's a trigonometric identity that's why we make that substitution now this cosines this squared here is going to take away this squared here okay so my integration actually in fact ends up to be pi over two right here negative pi over two right this is going to be cosine cube u and then cosine u again so then it's going to make it cosine to the four to the four to the four power right so cosine to the fourth power so this is a little bit cosine to the to the fourth power and then the u right so this one is quite simpler to work with so this is going to be it's not easy okay is you have to do a lot of things but you're just combining a little integration techniques okay to to solve this one okay this one is the same as cosine u, cosine square u squared, right? cosine to the 4 is cosine square squared.

05:08 I did that because i want to use something.

05:11 You want to have cosine to you.

05:15 It's the same as, okay, cosine square u minus one.

05:22 Okay, we've done this a lot of times.

05:26 Twice cosine square u minus one.

05:28 So if we solve for cosine square u, we're going to have cosine square u equals cosine to u plus 1 over 2, right? so we're going to replace this one by this guy right here.

05:46 Okay, so you see we're doing a lot of substitutions just to make our lives easier.

05:50 Okay, so by making a lot of substitutions, we can easily solve this integration.

05:57 So this one becomes a co -signituations.

06:00 2 u plus 1 over 2 all squared again right because there's a squared here right so you have the you right here and so see this two here is going to come out squared so i'm going to bring it out one of a four right here okay so two squared is four right then i have this one over two right here okay so when i have this one and then now i'm just going to have cosine 2 u plus 1 squared so it's just going to be cosine squared 2u plus twice, cosine 2u plus 1, right? i'm gonna have all the u right here, okay? so, well, it's been a long journey, but we're still going on.

06:47 Now, these two here are easily integrable, but the first one is not, we have to do another substitution, okay? so cosine 2u, plus 2 u is actually going to be uh cosine squared to u minus sine squared to u right and then sine squared sine squared to u is one minus cosine squared to u okay and so this is the same as twice cosine square to you minus one that is cosine for you okay so when you solve for this guy we're gonna have a cosine for you plus one over two equals cosine square to you so in place of this cosine square to you i'm gonna put this guys right here okay so my integration becomes like i said we're making a lot of substitutions that is the only way we can solve this integration problem because it is not as easy as it looks okay so cosine square to you is has been replaced by this one and then i have plus another integration right because i can make it i can i can do its term by term thing and then this is going to be twice, cosine to you, right? and then plus another integration of one, right? i've just splited the integration like that.

08:43 Let me come here to get better space.

08:46 Now with this one, i can do the integration now.

08:50 So this one is going to be cosine for you.

08:55 The integral of that one is going to be 1 over 4, right? sign for you and then the integration of this one is just going to be you do not forget there's a one half here because of this denominator okay and then it's is having the lower limit and then the upper limit right here and then here there's going to be a two here and then this integration is going to be also one or two okay and then sign sign to you okay and then also my upper and my lower limits right here and then finally this one is just going to be you and then also my upper and my lower limits hmm so i have this one okay there's no plus c because it is a definite integral okay so when you do your upper limit minus lower limit what do you have with this one, we do not change it back into the original integration because we changed the limit of integration.

10:17 Okay, so the answer is just going to be in the u form because we change the limits of integration.

10:23 Remember, the original limits of integration was negative one and positive one, but now they're negative power 2 and positive power 2...

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