2. INTRODUCTION
โฃ Powers of tangent functions go hand and hand with powers of secant
functions as shown below:
tan๐
๐ฅ sec๐
๐ฅ ๐๐ฅ
โฃ Differences in approach comes due to changes in the nature of the powers
of tangent and secant respectively.
โฃ ๐ and ๐ can either be even or old depending on the question given.
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3. EVEN POWER FOR SECANT
โฃ In one of the situations, secant will have an even power and tan will have
an odd power, that is, ๐ is even and ๐ is odd.
โฃ If this happens you write the integral as
tan๐ ๐ฅ sec๐โ2 ๐ฅ sec2 ๐ฅ ๐๐ฅ
โฃ Then, you have to write sec๐โ2
๐ฅ in terms as tan ๐ฅ using the identity
sec2 ๐ฅ = 1 + tan2 ๐ฅ. If ๐ โ 2 is greater than 2, you will express it as
power multiple of since it will be even. Letting ๐ข = tan ๐ฅ, the integrand
is simplified and sec2 ๐ฅ gets eliminated.
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4. EVEN POWER FOR SECANT
Example: Evaluate tan2
๐ฅ sec4
๐ฅ ๐๐ฅ.
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Firstly, we have to simplify ๐ก๐๐2 ๐ฅ ๐ ๐๐4 ๐ฅ using the rules provided.
tan2
๐ฅ sec4
๐ฅ = tan2
๐ฅ sec4โ2
๐ฅ sec2
๐ฅ = tan2
๐ฅ sec2
๐ฅ sec2
๐ฅ
But ๐ ๐๐2
๐ฅ = 1 + ๐ก๐๐2
๐ฅ, we substitute on one ๐ ๐๐2
๐ฅ and simplify.
= tan2
๐ฅ 1 + tan2
๐ฅ sec2
๐ฅ = (tan2
๐ฅ + tan4
๐ฅ) sec2
๐ฅ
This means that tan2
๐ฅ sec4
๐ฅ ๐๐ฅ = (tan2
๐ฅ + tan4
๐ฅ) sec2
๐ฅ ๐๐ฅ.
Using substitution method of integration, let ๐ข = tan ๐ฅ.
5. EVEN POWER FOR SECANT
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๐๐ข = sec2
๐ฅ ๐๐ฅ; ๐๐ฅ =
๐๐ข
sec2 ๐ฅ
Since the remaining function is ๐ข2 + ๐ข4 ๐ ๐๐2 ๐ฅ ๐๐ฅ, substitute ๐๐ฅ.
= (๐ข2 + ๐ข4) sec2 ๐ฅ ร
๐๐ข
sec2 ๐ฅ
= ๐ข2 + ๐ข4 ๐๐ข =
1
3
๐ข3 +
1
5
๐ข5 + ๐
After substituting ๐ข back, tan2
๐ฅ sec4
๐ฅ ๐๐ฅ =
1
3
tan3
๐ฅ +
1
5
tan5
๐ฅ + ๐.
6. EVEN POWER FOR SECANT
Below are the points we should take closer attention at.
โฃ When simplifying the function, on the first step we are having
tan๐ ๐ฅ sec๐โ2 ๐ฅ sec2 ๐ฅ ๐๐ฅ which is found by applying rules of indices.
โฃ The other thing we should see is that since ๐ is even, ๐ โ 2 will always
be even. Hence if it is greater than 2, the provided identity will always
work e.g. sec6โ2
๐ฅ sec2
๐ฅ = sec4
๐ฅ sec2
๐ฅ = 1 + tan2
๐ฅ 2
sec2
๐ฅ.
โฃ The idea of taking out sec2 ๐ฅ from sec๐ ๐ฅ is brought into account to
enable elimination through substituting the differentiation of tan ๐ฅ which
is sec2 ๐ฅ .
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7. โฃ Once the power secant is odd, the preceding method does not work. In this
case we consider another situation of having an odd power of tangent.
โฃ In this case you write the integral tan๐
๐ฅ sec๐
๐ฅ ๐๐ฅ where m is odd as shown
below:
tan๐โ1
๐ฅ sec๐โ1
๐ฅ tan ๐ฅ sec ๐ฅ ๐๐ฅ
โฃ Once this happen, we know that ๐ โ 1 is now even. That is, we will apply
sec2
๐ฅ = 1 + tan2
๐ฅ but this time to write tan ๐ฅ in terms of sec ๐ฅ. Then the
substitution of ๐ข = sec ๐ฅ simplifies the integrand.
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ODD POWERS OF TANGENT
8. Example: Evaluate tan3 ๐ฅ sec3 ๐ฅ ๐๐ฅ.
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ODD POWERS OF TANGENT
Firstly, we have to simplify ๐ก๐๐3 ๐ฅ ๐ ๐๐3 ๐ฅ using the rules provided.
tan3 ๐ฅ sec3 ๐ฅ = tan3โ1 ๐ฅ sec3โ1 ๐ฅ tan ๐ฅ sec ๐ฅ = tan2 ๐ฅ sec2 ๐ฅ tan ๐ฅ sec ๐ฅ
But ๐ ๐๐2
๐ฅ โ 1 = ๐ก๐๐2
๐ฅ, we substitute on tan2
๐ฅ and simplify.
= sec2 ๐ฅ โ 1 sec2 ๐ฅ tan ๐ฅ sec ๐ฅ = sec4 ๐ฅ โ sec2 ๐ฅ tan ๐ฅ sec ๐ฅ
Thus, tan3
๐ฅ sec3
๐ฅ ๐๐ฅ = sec4
๐ฅ โ sec2
๐ฅ tan ๐ฅ sec ๐ฅ ๐๐ฅ
Let ๐ข = sec ๐ฅ, ๐๐ข = tan ๐ฅ sec ๐ฅ ๐๐ฅ; ๐๐ฅ =
๐๐ข
tan ๐ฅ sec ๐ฅ
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ODD POWERS OF TANGENT
Since the remaining function is ๐ข4
โ ๐ข2
tan ๐ฅ sec ๐ฅ ๐๐ฅ, substitute ๐๐ฅ.
= ๐ข2 + ๐ข4 tan ๐ฅ sec ๐ฅ ร
๐๐ข
tan ๐ฅ sec ๐ฅ
= ๐ข4 โ ๐ข2 ๐๐ข
=
1
5
๐ข5 โ
1
3
๐ข3 + ๐
After substituting ๐ข back, tan3 ๐ฅ sec3 ๐ฅ ๐๐ฅ =
1
5
sec5 ๐ฅ โ
1
3
sec3 ๐ฅ + ๐.
10. โฃ If n is odd and m is even, this situation do not match any of the presented
method.
โฃ In such a case, the use other methods like integration by parts should be
used instead.
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EVEN POWERS OF TAN, ODD POWER OF SEC
11. Practice questions
1. Integrate tan3 ๐ฅ sec5 ๐ฅ with respect to ๐ฅ.
2. Evaluate the following integrals
a. tan2 ๐ฅ sec6 ๐ฅ ๐๐ฅ
b. cot3 ๐ฅ csc3 ๐ฅ ๐๐ฅ
c. tan5
๐ฅ sec ๐ฅ ๐๐ฅ
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POWERS OF TANGENT AND SECANT
13. โฃ Integrals involving trigonometric functions may grow complicated to an
extent that we apply the identities to simplify them. This is what we have
been so far.
โฃ But in some occasions, this is not always the story. We might tend to
encounter some integrands involving radicals which are seemingly
impossible to integrate.
โฃ For example, integrands like ๐2 โ ๐ฅ2, ๐2 + ๐ฅ2 and ๐ฅ2 โ ๐2 which
have no clear way to use in the integration process.
โฃ Therefore, the substitution of a specific trig function helps to eliminate
the radicals and usually simplifies the integrand.
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TRIGONOMETRIC SUBSITUTION
14. โฃ Before we dive deep into the process, you might wish to recall the
Pythagorean trig identity, cos2
๐ฅ + sin2
๐ฅ = 1 which gives birth to 1 +
tan2 ๐ฅ = sec2 ๐ฅ and cot2 ๐ฅ + 1 = csc2 ๐ฅ.
Table below shows specific trig substitution to match each given radical.
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Compiled by Kizio Makawa | Student - B.Sc Mathematics | University of Malawi | 0999978828 14
TRIGONOMETRIC SUBSITUTION
Expression Given Trig Substitution
๐2 โ ๐ฅ2 ๐ฅ = ๐sin ๐
๐2 + ๐ฅ2 ๐ฅ = ๐tan ๐
๐ฅ2 โ ๐2 ๐ฅ = ๐sec ๐
15. POINTS TO NOTE
โฃ When making substitutions we assume that ๐ is in the range of their
respective inverse functions.
โฃ If the radicals appear in the denominators, always add a condition
restricting the value of the denominator to never be equal to zero. For
example, if ๐2 โ ๐ฅ2 is a denominator always say โwhere ๐ฅ = ๐.
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TRIGONOMETRIC SUBSITUTION
16. Example: Evaluate
๐๐ฅ
๐ฅ2 16โ๐ฅ2
.
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TRIGONOMETRIC SUBSITUTION
Let ๐ = 16 = 4 and considering the radical, ๐ฅ = ๐sin ๐ = 4 sin ๐, thus;
๐๐ฅ
๐ฅ2 16 โ ๐ฅ2
=
๐๐ฅ
(4 sin ๐)2 16 โ 4 sin ๐ 2
=
1
16 sin2 ๐ 16(1 โ sin2 ๐)
๐๐ฅ
=
1
16 sin2 ๐ 16 cos2 ๐
๐๐ฅ since 1 โ sin2
๐ = cos2
๐
=
1
16 sin2 ๐ (4 cos ๐)
๐๐ฅ
Since ๐ฅ = 4 sin ๐ , ๐๐ฅ = 4๐๐๐ ๐๐๐. This will eliminate ๐๐ฅ and allow us to
integrate with respect to theta (๐).
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TRIGONOMETRIC SUBSITUTION
Eliminate dx and simplify.
=
1
16 sin2 ๐ (4 cos ๐)
ร 4 cos ๐ ๐๐ =
1
16 sin2 ๐
๐๐ =
1
16
csc2
๐ ๐๐
Therefore, it is simple to integrate csc2 ๐.
= โ
1
16
cot ๐ + ๐ถ
It is now necessary to get back to the original variable, ๐ฅ. Since ๐ฅ =
4 sin ๐ , sin ๐ =
๐ฅ
4
.
Therefore, ๐ = sinโ1 ๐ฅ
4
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Compiled by Kizio Makawa | Student - B.Sc Mathematics | University of Malawi | 0999978828 18
TRIGONOMETRIC SUBSITUTION
Using the triangle below,
Finally,
๐๐ฅ
๐ฅ2 16 โ ๐ฅ2
= โ
16 โ ๐ฅ2
16๐ฅ
+ ๐ถ
๐ฅ
4
42 โ ๐ฅ2
๐
cot ๐ =
1
tan ๐
=
๐๐๐๐๐๐๐๐ก
๐๐๐๐๐ ๐๐ก๐
Therefore cot ๐ =
16โ๐ฅ2
๐ฅ
19. POINTS TO NOTE
โฃ The idea mostly lies on selecting the correct substitution. Every other
thing is just mere algebra.
โฃ Always remember to change the integral back to the original variable, ๐ฅ.
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20. Practice questions
1. Evaluate the following integrals.
a.
1
4+๐ฅ2
๐๐ฅ
b.
๐ฅ2โ9
๐ฅ
๐๐ฅ
c.
1
๐ฅ4 ๐ฅ2โ3
๐๐ฅ
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End Of Session
Reach out for clarifications if necessary.