Trig Integrals Tan Sec

Summary

This video script delves into the integration of trigonometric functions, specifically focusing on combinations of tangent and secant. It outlines methods for tackling integrals with these functions, including substitution strategies for cases where the exponents are both odd or even, and the use of trigonometric identities. The script provides step-by-step examples, such as integrating tangent cubed times secant cubed, and demonstrates the application of reduction formulas and integration by parts for more complex scenarios, offering a comprehensive guide to solving advanced trigonometric integrals.

Takeaways

• 📚 The video discusses integrals involving trigonometric functions, specifically combinations of tangent and secant.
• 🔍 The relationship between tangent squared and secant squared is highlighted, where secant squared is 1 + tangent squared and vice versa.
• 📐 The integral of tangent to the m power and secant to the n power dx is tackled in different cases based on the parity of m and n.
• 🎲 The first case involves both m and n being odd integers, exemplified by the integral of tangent cubed x secant cubed x dx.
• 📉 A u-substitution technique is introduced where u is chosen to simplify the integral, such as letting u be cant x.
• 🔑 The identity tangent squared equals secant squared minus 1 is used to transform the integral into a more manageable form.
• 📈 The integral is simplified using algebraic manipulation and u-substitution, leading to an antiderivative in terms of u, which is then replaced back with the original trigonometric functions.
• 📝 The second case with both m and n being even integers is presented, such as the integral of tangent squared x secant squared x dx, resulting in a simpler antiderivative.
• 🤔 A third case where one exponent is odd and the other is even is explored, requiring a different approach and sometimes integration by parts.
• 🧩 The process of integration by parts is mentioned for cases where direct substitution is not feasible, such as when the powers of secant or tangent do not match up easily.
• 📚 The use of reduction formulas is alluded to for integrating odd powers of secant, which involves a recursive process of integration by parts.
• 📘 The final result of the integrals is expressed in terms of the original trigonometric functions, with constants of integration added.

Q & A

What is the relationship between tangent squared and secant squared?

Tangent squared (tan^2) is equal to secant squared minus one (sec^2 - 1), and secant squared is equal to one plus tangent squared (1 + tan^2).

How can you express the integral of tangent to the m power and secant to the n power in terms of u-substitution?

You can let u be a function of x that simplifies the integral, for example, u = tan(x) or u = sec(x), and then find du in terms of dx to perform the u-substitution.

What is the integral of tangent cubed x secant cubed x dx?

The integral is found by u-substitution, where u = sec(x), and then simplifying the expression to u^2 - 1, leading to the anti-derivative of u^4 - u^2, which is u^5/5 - u^3/3 + C, and replacing u back with sec(x).

How do you handle the integral when both exponents m and n are even?

When both exponents are even, you can perform a u-substitution where u is a function that simplifies the integral, such as u = tan(x), and then integrate u^2 du, which results in u^3/3 + C.

What is the integral of tangent to the 3rd power x secant to the 4th power x dx?

This integral is solved by recognizing that the even power of secant can be the u-substitution, resulting in an integral of u^3 * (1 + u^2) du, which simplifies to u^4/4 + u^6/6 + C after anti-derivatives.

What is the approach when one exponent is odd and the other is even?

When one exponent is odd and the other is even, you may need to use integration by parts or apply reduction formulas, especially when dealing with odd powers of secant.

How do you integrate secant cubed x dx?

Integration by parts is used, setting u = sec(x) and dv = sec^2(x)dx, which leads to a recursive integral that simplifies to sec(x)tan(x)/2 + (1/2)ln|sec(x) + tan(x)| + C after applying the reduction formula.

What is the significance of the identity tan^2(x) + 1 = sec^2(x) in trigonometric integrals?

This identity allows you to convert between tangent squared and secant squared, which is useful for simplifying integrals involving these trigonometric functions.

Can you provide an example of an integral that requires using both u-substitution and the identity tan^2(x) = sec^2(x) - 1?

An example is the integral of tangent cubed x secant squared x dx, where you first use u-substitution with u = sec(x) and then apply the identity to simplify the expression before integrating.

What is a reduction formula and how is it used in trigonometric integrals?

A reduction formula is a method used to simplify the integration of trigonometric functions with higher powers by reducing the power through integration by parts or other techniques, making the integral more manageable.

Outlines

00:00

📚 Trigonometric Integrals with Odd Powers of Tangent and Secant

This paragraph introduces the concept of integrating trigonometric functions involving combinations of tangent and secant raised to odd powers. The key identity used is that 1 + tangent squared is equal to secant squared, which allows for the expression of tangent squared in terms of secant squared and vice versa. The example given is the integral of tangent cubed times secant cubed of x with respect to x. The process involves setting u to be cant x, simplifying the integral using the derivative of u, and then substituting back to get the final answer in terms of secant x.

05:00

🔍 Integration of Trigonometric Functions with Even Powers

The second paragraph discusses the integration of trigonometric functions where both the powers of tangent and secant are even. The example provided is the integral of tangent squared x times secant squared x with respect to x. The method involves setting u to be tangent x, simplifying the integral to u squared with respect to u, and then integrating to obtain the result in terms of tangent cubed x over 3 plus a constant.

10:03

🔄 Odd and Even Power Combinations in Trigonometric Integrals

This paragraph explores the integration of trigonometric functions where one power is odd and the other is even. The example given is the integral of tangent cubed x times secant to the fourth power x with respect to x. The approach involves using substitution with u as secant squared x, simplifying the integral, and then integrating to find the result in terms of tangent x to the fourth power over four plus tangent x to the sixth power over six plus a constant.

📘 Advanced Trigonometric Integrals with Reduction Formulas

The final paragraph delves into more complex trigonometric integrals involving both odd and even powers, specifically focusing on the integral of secant to the fifth power x minus secant to the third power x with respect to x. The method involves using reduction formulas and integration by parts, with the identity tangent squared equals secant squared minus 1 being key. The process shows how to simplify the integral and eventually express the result in terms of secant x and its logarithmic function, plus a constant.

Keywords

💡Trigonometric Integrals

Trigonometric integrals are a type of calculus problem that involves integrating functions of the form involving sine, cosine, tangent, or secant. In the video, the focus is on integrals that combine tangent and secant functions, which are key to understanding the theme of the video.

💡Tangent

Tangent is a trigonometric function that relates the angle of a right triangle to the ratio of the opposite side to the adjacent side. In the script, tangent is raised to various powers and integrated, showing its importance in solving the integrals discussed.

💡Secant

Secant is the reciprocal of the cosine function and is used in trigonometry to relate the angle of a right triangle to the ratio of the hypotenuse to the adjacent side. The video script discusses integrals involving secant to various powers, demonstrating its role in the integrals.

💡U-Substitution

U-substitution is a technique used in calculus to simplify the process of integration. In the script, it is used to transform the integral of tangent and secant functions into a more manageable form by letting 'u' represent a related trigonometric expression.

💡Odd and Even Exponents

In the context of the video, odd and even exponents refer to the powers to which the tangent and secant functions are raised. The script discusses different cases based on whether these exponents are odd or even, which affects the approach to solving the integrals.

💡Reduction Formulas

Reduction formulas are used in calculus to simplify the integration of functions by reducing the power of the function in the integral. The script mentions the use of reduction formulas when dealing with integrals of secant to odd powers.

💡Integration by Parts

Integration by parts is a method in calculus for integrating a product of two functions. The script discusses using this method when the exponents of tangent and secant do not allow for a straightforward u-substitution, particularly when dealing with odd powers of secant.

💡Anti-Derivative

An anti-derivative is a function that, when differentiated, yields the original function. In the script, finding the anti-derivatives of expressions involving tangent and secant is a central part of solving the integrals.

💡Trigonometric Identities

Trigonometric identities are equations that hold true for all values of the variable and are used to simplify trigonometric expressions. The script uses identities such as 1 + tangent squared equals secant squared to transform the integrals into a form that can be more easily integrated.

💡Logarithmic Functions

Logarithmic functions are used in calculus to represent the inverse operation to exponentiation. In the script, the logarithm of secant plus tangent is part of the anti-derivative found for the integral of secant cubed.

💡Constant of Integration

The constant of integration is added to the antiderivative to account for the infinite number of original functions that could have the same derivative. The script mentions adding a constant of integration after finding the anti-derivatives of the integrals.

Highlights

Introduction to trigonometric integrals involving tangent and secant functions.

The relationship between 1 + tangent squared and secant squared is established.

Expression of tangent squared in terms of secant squared and vice versa.

Integral of tangent to the m power and secant to the n power dx is discussed in different cases.

Case analysis for when both m and n are odd integers.

Example integral of tangent cubed x secant cubed x dx is solved using substitution.

The use of trigonometric identities to simplify the integral.

Integration by substitution method for odd and even exponents of tangent and secant.

Solving the integral of tangent squared x secant squared x dx using u-substitution.

Case where both exponents are even, with an example of tangent to the 4th x secant to the 2nd x dx.

Integration by parts for cases where one exponent is odd and the other is even.

Example of integral of tangent cubed x secant to the 4th x dx using u-substitution and simplification.

The concept of reduction formulas in integration by parts for trigonometric functions.

Detailed walkthrough of integration by parts for secant cubed x dx.

Use of the identity tangent squared = secant squared - 1 in integration.

Final solution for the integral of secant cubed x dx involving logarithmic functions.

Application of the reduction formula to the integral of secant to the 5th x dx.

The complexity of cases where the odd power is secant and the need for integration by parts.

Conclusion on the process of solving trigonometric integrals with a combination of methods.

Transcripts

we're going to do trig integrals that

are combinations of tangent and secant

we can do this because 1 + tangent squar

an angle is secant squar so we

can express tangent squar in terms of

secant squar and vice versa so we know

that secant squ is just 1 + tangent

squar but we can say that tangent

squared is just secant s

-1 we're going to look at the integral

tangent to the m power secant to the N

power DX and we're going to do this in

cases our first

case is we're going to let M and N be

both odd

integers as an example we will do the in

integral of tangent cubed x secant cubed

x

DX in looking at this I'd like to have a

choice for you if I let U be

cant X then the derivative of U is

secant x tangent X DX that means I just

have to pull one from the tangent cubed

and one from the secant cubed

in doing this I will have my secant x

tangent X

DX right

there and that leaves me pulling one

from tangent with Tangent squar and one

from secant secant squar and that's

going to work fine because I can replace

a tangent squar or a secant

squared working this out I have tangent

squared secant squar and then I've got

my secant x tangent X

DX this is my du and the secants are my

U so now I have to change tangent squar

in terms of secant squar I do that from

the identity which says tangent squar is

secant squarus

1 this then becomes a u^2 minus1

here I can now write the U substitution

the integral becomes u^ 2

-1 * U 2ed and then I have

du we just distribute this

through and that gives us U to the 4th

minus U U ^ 2 du and now I'm ready to

take anti-derivative which is U 5th over

5 - U 3 over 3 plus a

constant but replacing U with the X

expression it is just secant x we've got

that right here so we write the answer

as secant x to 5 / 5 - secant x 3r / 3 +

a constant and that becomes our

answer our next case we are going to do

I'll call it case b where M and N are

both even are both even

exponents so I'm going to go to the next

page let's do as an example tangent s x

secant 2 x

DX my choice for you in this one would

be tangent cuz then the secant s DX is

my

[Music]

du this now becomes u 2 du so this is

the integral of u^2 du U which is U

cubed over 3 plus a

constant and it works out nicely this

becomes tangent cubed x over 3 plus a

constant

now we'll go to another example and our

other case is where one exponent is

odd and one is

even for this one as an example let's

look at the integral of tangent cubed

x secant to the 4 x DX

one way of thinking about this because I

have an even number of secant that's a

perfect choice for that to be my

du which means it came from mu being

tangent writing this out I have tangent

cubed x I've pulled out a secant squar

so I have secant SAR X and then I've got

the secant 2 x

DX this is my d U this is my U

cubed and that means this secant squ has

to become a tangent and it does because

it's 1 + tangent

squared I now have this as 1 + u^2 I'm

now ready to write it as the U

substitution it

is this becomes U cubed * 1 + u^ 2 and

then we've SC

du we just distribute the U Cube through

and this becomes the integral of U cubed

+ U to 5 duu and now very

straightforward anti-derivative U to 4

over 4 plus u 6/ 6 plus a

constant so this is just tangent X to 4/

4 + tangent X to 6 / 6 plus a constant

worked out very nicely we're now ready

for the example this first one is we had

secant was even but now we're going to

do one where tangent is even and secant

is

odd the example I'd like us to do would

be integral of

tangent SAR

X secant cubed x

DX this one I cannot um I can't let U be

tangent because when I take out a secant

squar I only have one secant I can't

replace it easily if I take out tangent

secant as du I have only one tangent so

this is going to involve integration by

parts so we're going to have to do by

Parts but before we do by parts parts we

are going to do a use an identity we're

going to change tangent squar to c^ 2 -

1 doing this this becomes the integral

of secant to the 5th xus secant to the

3r X

DX I'm going to do both of these

separately so this is secant to the 5th

x DX minus secant to the 3 x DX and it

turns out that this integral by

integration by parts is what's called a

reduction

formula and we could look it up but I'm

going to show you let's just do the

secan cubed x how you would do this if

you didn't see it before and you didn't

know the reduction formula I'm going to

go to the next

page I going to first do the integral of

secant cubed x DX and we are going to do

this as integration by parts to do that

I'm going to set u and DV and then

afterward I have du and and v i look at

SEC cubed and I realize the only thing

that DV could be would be c^ 2 x DX cuz

that's something that I know the

anti-derivative of

so DV would be secant 2 x

DX and U would be secant

x du the derivative of secant is just

secant x tangent X

DX but the anti-derivative of secant

squar is tangent

X working this we're going to go it's

this minus the product of

this we get that the integral of secant

cubed x DX is crossing this it's secant

x * tangent x

minus the integral of secant x tangent 2

x

DX at this point I'm not sure what to do

so I am going to replace tangent

with c^ 2

-1 looking at this look what I can write

I can say secant cubed x DX is secant x

tangent x

minus the integral of secant cubed x

DX but then I had minus a minus with the

secant is plus the integral of secant X

DX but this can get added to the other

side in doing that what we get is we've

got 2 SEC cubed x DX is secant x tangent

X and not minus but Plus

plus the integral of secant x DX and

this integral this last integral here we

have to memorize what that answer

is continuing we get two cant the

integral of secant Cub DX is just secant

x tangent X plus the anti-derivative

that we've memorized of secant is the

log of secant x plus tangent X plus a

constant and sure enough the last step

is we will divide by

two dividing by

two we can put a half here and we don't

have to divide the constant by two it's

stays therefore we get that the integral

of secant cubed x DX is just secant x

tangent x/ 2 and this log over 2 or I

write it as 12 the log of secant X+

tangent X plus a constant and we could

do the same with the integral of secant

to the 5th so going back here this is

the trickiest of all the

cases when in this particular

case when the odd one is the secant we

have to do by part

and that's going to involve changing to

cants and odd powers of

cant or what we call reduction formulas

which involve integration by parts and

when we're finished we're not completely

finished we'd have to do the same thing

for the integral of secant to the 5th