Trig Integrals Tan Sec
Summary
TLDRThis 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
📚 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.
🔍 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.
🔄 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.
Mindmap
Keywords
💡Trigonometric Integrals
💡Tangent
💡Secant
💡U-Substitution
💡Odd and Even Exponents
💡Reduction Formulas
💡Integration by Parts
💡Anti-Derivative
💡Trigonometric Identities
💡Logarithmic Functions
💡Constant of Integration
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
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