Evaluate the integrals ∫0^π/ 3(cosx+secx)^2 d x

Name: Problem 26, Chapter 5: Thomas Calculus
Uploaded: 2019-05-12T22:41:43-08:00
Duration: 17 min 28 s
Channel: Nathalie Luna
Description: Evaluate the integrals ∫0^π/ 3(cosx+secx)^2 d x

Key Concepts

Definite Integration

Definite integration involves computing the value of an integral over a specific interval using the Fundamental Theorem of Calculus. This process requires evaluating the antiderivatives at the boundary values of the interval.

Expansion of Algebraic Expressions

This concept involves using algebraic techniques to expand and simplify expressions, such as squaring a sum. It is important for breaking down complex integrals into simpler components that can be individually integrated.

Trigonometric Identities

Trigonometric identities, such as the relationship between cosine and secant, are essential for rewriting and simplifying trigonometric expressions. They allow the conversion of complex expressions into forms that are more straightforward to integrate.

Integration of Trigonometric Functions

This concept covers the methods and formulas used for integrating functions involving trigonometric expressions. It includes identifying standard integrals and applying specific techniques to handle squared trigonometric functions.

Transcript

00:01 So in this problem, we are asked to evaluate the integral from 0 to pi over 3 of this binomial squared with respect to x.

00:15 So the first thing that we're going to do is we are going to rewrite the integral as several terms as opposed to this square.

00:26 So the so we are so blah.

00:31 So the first thing is to expand this binomial squared.

00:36 And to do that, we, oh, it's out here.

00:41 So to do that, we do not want to, we don't want to sort of distribute this exponent.

00:56 So what i mean is if we have a binomial that is squared, that is not equal to first term squared plus second term squared.

01:12 Instead, this is equal to first term squared plus two times first term times second term, plus second term, plus second term, so this is what we are going to do for the integrand.

01:38 So we have, we're not integrating just yet, so we need to keep this integral from 0 to pi over 3, cosine first term squared, plus 2 times the first term, cosine x times the second term, that is secan, of x and finally we add the second term squared that is ccan squared of x so this is the integral and we specify with respect to which variable we're integrating so now we're working with a more manageable integral so first term we have cosine squared which we don't have an anti -derivative form, but we can rewrite this term using the double angle formula.

02:56 So in this formula, we have this cosine square x.

03:03 So what we're going to do is we're going to solve for cosine square.

03:10 And what we get, that is going to be our, that's what we're going to substitute in the integral.

03:20 So first we get rid of this one.

03:23 So we do it on the other side.

03:26 This cancels out.

03:30 So we have cosine of 2x equal, oops, plus 1, is equal to 2 times cosine squared.

03:48 Next we get rid of this 2.

03:50 So we divide by 2 here, we divide by 2 on the other side, these two cancel out and we have cosine squared of x is equal to, we can split this.

04:06 We have one half of cosine of 2x plus 1 over 2.

04:19 So now we have this other expression for cosine square of x, and this is definitely more manageable for, or this is better.

04:30 To integrate.

04:34 This we are going to substitute with one half cosine two of x plus one half.

04:45 Now this term we can simplify it and we are going to work on that over here.

04:55 So cosine we leave us is cosine of x.

05:01 Now secant by definition is 1 over cosine of x so voila these two cancel out this is in the numerator this is the denominator so this is actually equal to one so that simplifies by a lot and so this is this is equal to one now as for ccm squared i don't want to do any with it because we know that the derivative of tangent of x with respect to x is second of x sqan squared so we have here an antiderivative so we don't need to touch this so adding in this green stuff here remember we're not deriving it i'm sorry we're not integrating yet so we keep this here so the first term is 1 .5 cosine of 2x plus 1 half.

06:30 Next is 2 times 1.

06:34 That's just 1.

06:35 I'm sorry, that's just 2.

06:37 And we said we were not going to do anything with 2kn squared.

06:44 So this is what we did here.

06:48 Now, are we ready to integrate? not just yet because i want to know.

06:54 How to derive this.

06:56 I mean it's certainly easier than cosine squared, but we have cosine and then we have in the argument we have another another function which would be 2 of x and how do we deal with that we have it's not the power rule so we have a kind of reverse chain roll so if we have this integral of f and instead of x we have this new function called g of x and this is multiplied by g prime of x so that's the derivative of the inside inside function and we are integrating with respect to x then if we have this setup we can integrate on f treating gfx as a new constant.

08:02 So we have this big f of gfx.

08:09 This is the new or of the new variable.

08:15 So in our case we have, in this case, we have ffx equal to cosine of equal to cosine of x.

08:34 And we have gfx to be equal.

08:39 To 2 of x now we don't have the g prime right because what we have is one half but but g prime of x is actually 2 so it's not it's not one and a half which is what we do have but since it's a constant it's very easy to get this set up so how do we do that we just so think of this this is we can think of this as having a denominator of 1.

09:19 So we can multiply by 2 here, and we do the same here, and we're allowed to do that.

09:28 So with this, this becomes a sort of, this one and a half, and this one half become a coefficient, and then this cosine 2 of x times 2.

09:42 That is the setup for this formula...

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