Section 6.2 - Trig integrals and substitution - Part 2
Summary
TLDRThis educational video script focuses on integral calculus, specifically integrals involving powers of tangent and secant functions. It outlines strategies for solving integrals when the power of tangent is odd or the power of secant is even, and when both conditions are met simultaneously. The script provides step-by-step examples, demonstrating how to simplify integral expressions using trigonometric identities and substitution methods. It emphasizes the importance of recognizing when the power of tangent is odd, which allows for the use of specific identities to simplify the integral. The tutorial is designed to help viewers understand and tackle complex integrals in calculus.
Takeaways
- 📚 The video discusses strategies for integrating functions involving powers of tangent and secant.
- 🔑 There are two straightforward cases: when the power of tangent is odd or the power of secant is even.
- ❌ If the power of tangent is even and the power of secant is odd, a generic strategy doesn't exist, and a case-by-case approach is needed.
- 🤔 The strategy involves pulling out one tangent or secant from the integral to simplify the expression.
- 🆚 For an odd power of tangent, after pulling out one tangent, the remaining power is even, allowing the use of a specific trigonometric identity.
- 🆓 The identity used is tangent squared equals secant squared minus one, which helps rewrite the integral in terms of secant.
- 🔄 A u-substitution is performed where u is secant, and du is secant tangent, simplifying the integration process.
- 📉 The integral is then integrated with respect to u, and the results are replaced back in terms of the original variable.
- 🔠 For an even power of secant, the strategy is to pull out secant squared and use the identity secant squared equals tangent squared plus one.
- 🔢 The final answer is expressed in terms of the original variable, with the integrated result adjusted for the pulled-out secant or tangent.
Q & A
What are the two cases where it is easy to solve integrals involving powers of tangent and secant?
-It is easy to solve when the power of tangent is odd or the power of secant is even. If both conditions are met, either method can be used to find the correct answer.
Why is it helpful when the power of tangent is odd in solving integrals?
-When the power of tangent is odd, you can pull out one tangent and then use trigonometric identities to simplify the remaining even power of tangent, making it easier to solve.
What is the first step when dealing with an integral where the power of tangent is odd?
-The first step is to pull out one tangent and one secant from the integral. This allows you to work with an even power of tangent, which can be simplified using a trigonometric identity.
What trigonometric identity is used to simplify the integral after pulling out one tangent and secant?
-The identity used is tangent^2(x) = secant^2(x) - 1, which allows you to rewrite the remaining even power of tangent in terms of secant.
Why is a substitution method used in these integrals, and what is the common substitution?
-A substitution method is used to simplify the integral into a form that is easier to integrate. The common substitution is u = secant(x), where the derivative of secant is secant(x) * tangent(x), helping to reduce the complexity.
How do you rewrite the integral after the substitution u = secant(x)?
-After substitution, the integral becomes a polynomial in terms of u, which can be expanded and integrated using basic rules of integration.
What happens if the power of tangent is even and the power of secant is odd?
-If the power of tangent is even and the power of secant is odd, the integral becomes more difficult, and no generic strategy can be applied. Each case must be handled individually.
What is the strategy when the power of secant is even?
-When the power of secant is even, the strategy is to pull out a factor of secant^2 and then use the identity secant^2(x) = 1 + tangent^2(x) to express the integral in terms of tangent.
How do you handle an integral where both tangent and secant have powers greater than 2?
-In such cases, you first pull out secant^2 and use the identity secant^2(x) = 1 + tangent^2(x) to rewrite everything in terms of tangent, then proceed with substitution.
What is the final form of the integral after solving for an odd power of tangent and an even power of secant?
-The final form of the integral is a polynomial in secant(x), which can be integrated and expressed in terms of secant(x) to a power, plus a constant (C).
Outlines
📚 Understanding Integrals Involving Tangent and Secant Powers
The video begins by introducing integrals that involve powers of tangent and secant. The key strategies are discussed for cases where the power of tangent is odd or the power of secant is even, and how to handle situations where both conditions apply. In cases where the power of tangent is odd, the method involves pulling out one tangent and one secant, rewriting the integral, and using trigonometric identities to simplify it. The instructor explains how to rewrite the integral in terms of secant using the identity for tangent squared and secant squared, and then applies u-substitution to solve the integral.
✏️ Applying U-Substitution and Completing the Integral
This section covers the step-by-step process of applying u-substitution to solve the integral. The instructor walks through distributing terms to integrate powers of u, solving for u, and then replacing u back with secant. The importance of the odd power of tangent is emphasized, as it allows the use of the identity for tangent squared. After completing the integration, the instructor showcases the final answer in terms of secant raised to different powers, highlighting the significance of the odd power of tangent for the method.
🔄 Example of Even Powers of Secant and Tangent
The final paragraph provides a contrasting example where the power of tangent is even, demonstrating a different approach. Here, the strategy focuses on pulling out secant squared and using the identity that relates secant squared to tangent. The integral is then rewritten entirely in terms of tangent, and the same u-substitution method is applied. The instructor solves the integral step-by-step, foiling terms before integrating, and concludes by replacing u with tangent to show the final result. The method's effectiveness for even powers of secant is emphasized.
Mindmap
Keywords
💡Tangent
💡Secant
💡Power of tangent (odd)
💡Power of secant (even)
💡Trigonometric identity
💡U-substitution
💡Derivative
💡Integral
💡Distribution
💡Trigonometric substitution
Highlights
Two cases are highlighted for solving integrals involving powers of tangent and secant: when the power of tangent is odd, or the power of secant is even.
If the power of tangent is odd, one can pull out one tangent and one secant to simplify the integral.
The integral can be rewritten with the remaining terms as a function of the derivative of secant and tangent.
The identity involving tangent squared and secant squared is used to transform the integral into a more manageable form.
A U-substitution is performed with U as secant, allowing for an easier integration.
The integral of a function involving secant and tangent can be solved by expressing it in terms of secant alone.
The final answer is expressed in terms of secant to a certain power, showcasing the successful integration.
When the power of tangent is odd, the integral can be simplified by using the identity for tangent squared.
An example is provided where the power of secant is even, requiring a different approach.
In the case of an even power of secant, one can pull out a secant squared to simplify the integral.
The identity secant squared is equal to tangent of s plus 1 is used to rewrite the integral in terms of tangent.
A U-substitution with U as tangent is performed to integrate the function in terms of tangent.
The integral is solved by expressing it in terms of tangent and then integrating.
The final answer is expressed in terms of tangent to a certain power, completing the integration process.
The strategy for dealing with powers of tangent times powers of secant is outlined, emphasizing the importance of the power's parity.
Transcripts
[Music]
so now let's look at some integrals
involving powers of tangent and
secant so there are two cases in which
it's easy to do something and those
cases are if the power of tangent is odd
or the power of secant is
even okay or if it's both at the same
time like if this were even and this
were odd you could use either method and
again you're answer would look a little
different but they'd both be
correct if this is
even and this one is
odd no luck we don't know what to do
then you have to go on a Case by case
basis and see what you can do
okay so let's just do one example of
each one of those and we'll kind of put
the strategy on the right
here okay so let's first deal with the
um the uh possibility that the m is odd
so the power of tangent is odd okay so
what you're going to do is you're going
to pull
out one tangent so here I wrote the
variable as tangent of R just to get you
practicing the feeling of having
different variables so pull out one
tangent of whatever your variable is
here I'll WR R and one also
secant okay and I'm going to put them
together at the end
so this is what I mean so I'm going to
rewrite this integral I'm going to take
out one of the tangents here so I'll be
left with Tangent
squar and I'll have an extra tangent at
the
end and I'm going to pull out one of the
secants here so I'll have secant to the
5th sorry secant to the
4th and I'll just write this one
here so do you agree that that this is
equal to
this okay CU I have tangent * tangent
that's tangent cubed and secant to 4th
tangent squ time tangent is tangent
cubed and secant to 4th * secant gives
me secant to the 5th remember this is
not multiplication okay this is one
object object tangent of R cubed here's
a multiplication and this is another
object secant of R to the 5th okay why
was that helpful to me
because
this is going to play the role of my du
if that's my du what do you think my U
is going to be it's going to be
something whose derivative is secant
tangent and if you go back to our
first
page that was this so I'm going to have
a u being
secant that
means that I want everything left to be
in terms of secant
this is already in terms of cant so this
is good and this I need to
replace okay so the second part of my
strategy is use the
identity okay so let's go sorry to make
you a little nauseous here let's use the
identity involving tangent and secant so
this one so I'm going to replace
everything to do with Tangent squar by
secant 2us 1 so if
tangent + 1 is secant
squ I can subtract the one from both
sides and get this okay so I'm going to
use the
identity that tangent
squared in my case I have R is secant s
rus1 to rewrite my
integral in terms
of
secant okay so let's do it so this is
the integral of^ 2 R -1 * 4 of R time
secant R tangent
R doesn't really matter whether I wrote
tangent time secant or secant time
tangent I just wrote it in this
particular order because it's familiar
to us as a derivative okay if this had
been tangent to the
6th then I would first have written this
tangent squared
cubed and then replaced by this and I
would have had a cube
here okay so I want to see tangent
squared in my
integral okay and now I do my U
substitution I take U as
secant so du is secant
tangent everything's perfectly set up
for
me so this becomes the integral
of u^2 - 1
* U 4
du first I have to distribute to be able
to integrate U ^2 * U 4 is U the 6 -1 *
U 4th is U
4th I integrate and then I replace the U
I'll do it in two steps but you guys I'm
sure by now are already good at
this okay so I would integrate
and then in the end replace
you so then my final answer would be 17
U was secant this would be secant to the
7 in my case my variable was called R
minus
1 or I could write secant to the 5th
over 5 or 1 secant to the 5th plus C
okay so this would be my final answer
here
okay okay so that's the example if m is
odd if the power of tangent is odd it
mattered that it was odd because when I
took one out I had an even power of
tangent left which means I could use
this identity and even power can be
written as tangent squared to some power
this case it was just tangent squared
itself which allowed me to use this
identity so that's why it mattered that
my M was
odd okay so let's do one more example
like this
now I called my variable
s and you see this is not odd so if I
took out one of my tangents
here I'd be left with Tangent cubed and
then there's no way to use that identity
in terms of tangent squar of anything
okay so that's a nogo so instead I'm
going to use the fact that this is even
this is even and this were odd then I'm
in trouble hard integral okay then you
have to try to find some other way there
there's no generic strategy but if this
is even
now um my generic strategy is
to use the fact that I have an even
power of
secant so my strategy is to pull
out a secant squared of
something okay so let's do that here so
so I'm going to have tangent to
4th secant to 4th is the same as secant
s * secant
squ and then my idea is this is the
derivative of something this is the
derivative of
tangent okay so my second
part is to use the identity the same
identity so what would be good for me is
if everything here were written in terms
of tangent because that would be my U
and this would be my du so that's my
problem guy left that this is not in
terms of
tangent okay so I'm going to use the
identity that secant squ of
s
is tangent s of s +
1 okay that's the identity I'm going to
use and so in this
case so I'm going to use this to rewrite
the rest of my
integral in terms of
tangent okay so let's do that
here it's kind of fun somehow in a very
geeky sort of way okay so secant squ is
the same as tangent s s +
1 this I keep the same this du I keep
the
same and then I use a u
substitution U is
tangent so my du is secant
squ and I
integrate so this is U to 4th this is
u^2 + 1 this is
Du I have to foil before I integrate
this is U 6 Plus U 4 du
this is 17th U 7th + 1/5 U 5th if I make
a mistake along here you can check
me so and then finally after I integrate
I replace
you 17th my U was tangent I go back to
my original variable which was s in this
case tangent to the 5 of s plus c
okay okay so that's my strategy for
dealing with powers of tangent time
powers of secant
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