Integrals involving powers of tan and trig substitution.pptx

INTRODUCTION
‣ Powers of tangent functions go hand and hand with powers of secant
functions as shown below:
tan𝑚
𝑥 sec𝑛
𝑥 𝑑𝑥
‣ Differences in approach comes due to changes in the nature of the powers
of tangent and secant respectively.
‣ 𝒎 and 𝒏 can either be even or old depending on the question given.
Thursday, 28 March 2024
Compiled by Kizio Makawa | Student - B.Sc Mathematics | University of Malawi | 0999978828 2

EVEN POWER FOR SECANT
‣ In one of the situations, secant will have an even power and tan will have
an odd power, that is, 𝑛 is even and 𝑚 is odd.
‣ If this happens you write the integral as
tan𝑚 𝑥 sec𝑛−2 𝑥 sec2 𝑥 𝑑𝑥
‣ Then, you have to write sec𝑛−2
𝑥 in terms as tan 𝑥 using the identity
sec2 𝑥 = 1 + tan2 𝑥. If 𝑛 − 2 is greater than 2, you will express it as
power multiple of since it will be even. Letting 𝑢 = tan 𝑥, the integrand
is simplified and sec2 𝑥 gets eliminated.

Example: Evaluate tan2
𝑥 sec4
𝑥 𝑑𝑥.
Firstly, we have to simplify 𝑡𝑎𝑛2 𝑥 𝑠𝑒𝑐4 𝑥 using the rules provided.
tan2
𝑥 sec4
𝑥 = tan2
𝑥 sec4−2
𝑥 sec2
𝑥 = tan2
𝑥 sec2
𝑥 sec2
𝑥
But 𝑠𝑒𝑐2
𝑥 = 1 + 𝑡𝑎𝑛2
𝑥, we substitute on one 𝑠𝑒𝑐2
𝑥 and simplify.
= tan2
𝑥 1 + tan2
𝑥 sec2
𝑥 = (tan2
𝑥 + tan4
𝑥) sec2
𝑥
This means that tan2
𝑥 sec4
𝑥 𝑑𝑥 = (tan2
𝑥 + tan4
𝑥) sec2
𝑥 𝑑𝑥.
Using substitution method of integration, let 𝑢 = tan 𝑥.

𝑑𝑢 = sec2
𝑥 𝑑𝑥; 𝑑𝑥 =
𝑑𝑢
sec2 𝑥
Since the remaining function is 𝑢2 + 𝑢4 𝑠𝑒𝑐2 𝑥 𝑑𝑥, substitute 𝑑𝑥.
= (𝑢2 + 𝑢4) sec2 𝑥 ×
𝑑𝑢
sec2 𝑥
= 𝑢2 + 𝑢4 𝑑𝑢 =
1
3
𝑢3 +
1
5
𝑢5 + 𝑐
After substituting 𝑢 back, tan2
𝑥 sec4
𝑥 𝑑𝑥 =
1
3
tan3
𝑥 +
1
5
tan5
𝑥 + 𝑐.

Below are the points we should take closer attention at.
‣ When simplifying the function, on the first step we are having
tan𝑚 𝑥 sec𝑛−2 𝑥 sec2 𝑥 𝑑𝑥 which is found by applying rules of indices.
‣ The other thing we should see is that since 𝑛 is even, 𝑛 − 2 will always
be even. Hence if it is greater than 2, the provided identity will always
work e.g. sec6−2
𝑥 sec2
𝑥 = sec4
𝑥 sec2
𝑥 = 1 + tan2
𝑥 2
sec2
𝑥.
‣ The idea of taking out sec2 𝑥 from sec𝑚 𝑥 is brought into account to
enable elimination through substituting the differentiation of tan 𝑥 which
is sec2 𝑥 .

‣ 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.
‣ In this case you write the integral tan𝑚
𝑥 sec𝑛
𝑥 𝑑𝑥 where m is odd as shown
below:
tan𝑚−1
𝑥 sec𝑛−1
𝑥 tan 𝑥 sec 𝑥 𝑑𝑥
‣ Once this happen, we know that 𝑚 − 1 is now even. That is, we will apply
sec2
𝑥 = 1 + tan2
𝑥 but this time to write tan 𝑥 in terms of sec 𝑥. Then the
substitution of 𝑢 = sec 𝑥 simplifies the integrand.
ODD POWERS OF TANGENT

Example: Evaluate tan3 𝑥 sec3 𝑥 𝑑𝑥.
Firstly, we have to simplify 𝑡𝑎𝑛3 𝑥 𝑠𝑒𝑐3 𝑥 using the rules provided.
tan3 𝑥 sec3 𝑥 = tan3−1 𝑥 sec3−1 𝑥 tan 𝑥 sec 𝑥 = tan2 𝑥 sec2 𝑥 tan 𝑥 sec 𝑥
But 𝑠𝑒𝑐2
𝑥 − 1 = 𝑡𝑎𝑛2
𝑥, we substitute on tan2
𝑥 and simplify.
= sec2 𝑥 − 1 sec2 𝑥 tan 𝑥 sec 𝑥 = sec4 𝑥 − sec2 𝑥 tan 𝑥 sec 𝑥
Thus, tan3
𝑥 sec3
𝑥 𝑑𝑥 = sec4
𝑥 − sec2
𝑥 tan 𝑥 sec 𝑥 𝑑𝑥
Let 𝑢 = sec 𝑥, 𝑑𝑢 = tan 𝑥 sec 𝑥 𝑑𝑥; 𝑑𝑥 =
𝑑𝑢
tan 𝑥 sec 𝑥

Since the remaining function is 𝑢4
− 𝑢2
tan 𝑥 sec 𝑥 𝑑𝑥, substitute 𝑑𝑥.
= 𝑢2 + 𝑢4 tan 𝑥 sec 𝑥 ×
𝑑𝑢
tan 𝑥 sec 𝑥
= 𝑢4 − 𝑢2 𝑑𝑢
=
1
5
𝑢5 −
1
3
𝑢3 + 𝑐
After substituting 𝑢 back, tan3 𝑥 sec3 𝑥 𝑑𝑥 =
1
5
sec5 𝑥 −
1
3
sec3 𝑥 + 𝑐.

‣ If n is odd and m is even, this situation do not match any of the presented
method.
‣ In such a case, the use other methods like integration by parts should be
used instead.
EVEN POWERS OF TAN, ODD POWER OF SEC

Practice questions
1. Integrate tan3 𝑥 sec5 𝑥 with respect to 𝑥.
2. Evaluate the following integrals
a. tan2 𝑥 sec6 𝑥 𝑑𝑥
b. cot3 𝑥 csc3 𝑥 𝑑𝑥
c. tan5
𝑥 sec 𝑥 𝑑𝑥
POWERS OF TANGENT AND SECANT

‣ 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.
‣ 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.
‣ For example, integrands like 𝑎2 − 𝑥2, 𝑎2 + 𝑥2 and 𝑥2 − 𝑎2 which
have no clear way to use in the integration process.
‣ Therefore, the substitution of a specific trig function helps to eliminate
the radicals and usually simplifies the integrand.
TRIGONOMETRIC SUBSITUTION

‣ Before we dive deep into the process, you might wish to recall the
Pythagorean trig identity, cos2
𝑥 + sin2
𝑥 = 1 which gives birth to 1 +
tan2 𝑥 = sec2 𝑥 and cot2 𝑥 + 1 = csc2 𝑥.
Table below shows specific trig substitution to match each given radical.
Expression Given Trig Substitution
𝑎2 − 𝑥2 𝑥 = 𝑎sin 𝜃
𝑎2 + 𝑥2 𝑥 = 𝑎tan 𝜃
𝑥2 − 𝑎2 𝑥 = 𝑎sec 𝜃

POINTS TO NOTE
‣ When making substitutions we assume that 𝜃 is in the range of their
respective inverse functions.
‣ 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 𝑎2 − 𝑥2 is a denominator always say “where 𝑥 = 𝑎.

Example: Evaluate
𝑑𝑥
𝑥2 16−𝑥2
.
Let 𝑎 = 16 = 4 and considering the radical, 𝑥 = 𝑎sin 𝜃 = 4 sin 𝜃, thus;
𝑑𝑥
𝑥2 16 − 𝑥2
=
𝑑𝑥
(4 sin 𝜃)2 16 − 4 sin 𝜃 2
=
1
16 sin2 𝜃 16(1 − sin2 𝜃)
𝑑𝑥
=
1
16 sin2 𝜃 16 cos2 𝜃
𝑑𝑥 since 1 − sin2
𝜃 = cos2
𝜃
=
1
16 sin2 𝜃 (4 cos 𝜃)
𝑑𝑥
Since 𝑥 = 4 sin 𝜃 , 𝑑𝑥 = 4𝑐𝑜𝑠𝜃𝑑𝜃. This will eliminate 𝑑𝑥 and allow us to
integrate with respect to theta (𝜃).

Eliminate dx and simplify.
=
1
16 sin2 𝜃 (4 cos 𝜃)
× 4 cos 𝜃 𝑑𝜃 =
1
16 sin2 𝜃
𝑑𝜃 =
1
16
csc2
𝜃 𝑑𝜃
Therefore, it is simple to integrate csc2 𝜃.
= −
1
16
cot 𝜃 + 𝐶
It is now necessary to get back to the original variable, 𝑥. Since 𝑥 =
4 sin 𝜃 , sin 𝜃 =
𝑥
4
.
Therefore, 𝜃 = sin−1 𝑥
4

Using the triangle below,
Finally,
𝑑𝑥
𝑥2 16 − 𝑥2
= −
16 − 𝑥2
16𝑥
+ 𝐶
𝑥
4
42 − 𝑥2
𝜃
cot 𝜃 =
1
tan 𝜃
=
𝑎𝑑𝑖𝑎𝑐𝑒𝑛𝑡
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
Therefore cot 𝜃 =
16−𝑥2
𝑥

POINTS TO NOTE
‣ The idea mostly lies on selecting the correct substitution. Every other
thing is just mere algebra.
‣ Always remember to change the integral back to the original variable, 𝑥.

Practice questions
1. Evaluate the following integrals.
a.
1
4+𝑥2
𝑑𝑥
b.
𝑥2−9
𝑥
𝑑𝑥
c.
1
𝑥4 𝑥2−3
𝑑𝑥
End Of Session
Reach out for clarifications if necessary.

Integrals involving powers of tan and trig substitution.pptx

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Integrals involving powers of tan and trig substitution.pptx