Integral Calculus is a branch of mathematics that helps us find the total or whole by adding up many small parts. It is the opposite of differentiation. When we use integral calculus, we are often trying to find areas, volumes, or the total value of a changing quantity.

In simple terms, if a function shows how something changes (like speed, height, or temperature), integration helps us find the total of that change over time or space.

Many types of mathematical functions can be integrated, including trigonometric functions like sin(x), cos(x), and tan(x). These are commonly used in physics, engineering, and other fields.

What is the Integral of Tanx?

The integral of tan(x) is written as −ln|cos(x)| + C or ln|sec(x)| + C.

The function tan(x) is continuous everywhere except at points like π/2, 3π/2, 5π/2, etc., where it is not defined.

So, the domain of tan(x) includes all real numbers except those odd multiples of π/2.

Tan is a trigonometric function that relates the two sides of a right angled triangle to an angle. Tan or Tangent is a mathematical ratio of the opposite sides and adjacent sides. Thus,

\(tan(\theta) = \frac{\text{opposite side}}{\text{adjacent side}}\)

We can integrate tanx to get value as

\({\int}tan(x)dx=−ln(cos(x))+C\)

Continue reading about second order derivatives.

To find the Integral of Tanx we use integration by substitution method. Let’s learn how we can integrate Tanx. 

The easiest way to integrate tan(x) is to recall that

\(tan(x) = \frac{sin(x)}{cos(x)}\)

so

\({\int}tan(x)dx = {\int}\frac{sin(x)}{cos(x)}dx\)

We use the substitution method to integrate this function

We substitute u = cos(x), du = -sinx dx.

\({\int}\frac{sin(x)}{cos(x)}dx = {\int}\frac{1}{u}du\)

\({\int}\frac{sin(x)}{cos(x)}dx = -log(u)+c\) 

Resubstituting cos(x) = u, we get;

\({\int}\frac{sin(x)}{cos(x)}dx = -log(cos(x))+c\)

\({\int}tan(x)dx = -log(cos(x))+c\)

\({\int}tan(x)dx = log(cos(x)^{-1})+c\)

\({\int}tan(x)dx = log(\frac{1}{cos(x)})+c\)

\({\int}tan(x)dx = log(sec(x))+c\)

Continue reading more about First Principles of Derivative here.

Definite Integration of Tanx

A definite integral is the area under a curve between two fixed limits. The process of finding integrals is called integration. Definite integrals are used when the limits are defined to generate a unique value. 

We apply the formula of definite integrals

\({\int}^{b}_{a}f(x)dx = f(b) - f(a)\)

Example: Let’s integrate Tanx between \([-\frac{\pi}{4}, \frac{\pi}{4}]\)

Solution:

\(tan(x) = \frac{sin(x)}{cos(x)}\)

so

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = {\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{sin(x)}{cos(x)}dx\)

We use the substitution method to integrate this function

We substitute u = cos(x), du = -sinx dx.

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{sin(x)}{cos(x)}dx={\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{1}{u}du\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{sin(x)}{cos(x)}dx=[-log(u)+c]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\) 

Resubstituting cos(x) = u, we get;

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\frac{sin(x)}{cos(x)}dx=[-log(cos(x))+c]]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [-log(cos(x))+c]]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [log(cos(x)^{-1})+c]]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx=[log(\frac{1}{cos(x)})+c]]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [log(sec(x))+c]^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [log(sec(\frac{\pi}{4}))+c]-[log(sec(-\frac{\pi}{4}))+c]\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [log(sec(\frac{\pi}{4}))+c+log(sec(\frac{\pi}{4}))-c]\)

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [2log(sec(\frac{\pi}{4}))]\)

We know that, 

\(sec(\frac{\pi}{4})=\sqrt{2}\)

Therefore, 

\({\int}^{\frac{\pi}{4}}_{-\frac{\pi}{4}}tan(x)dx = [2\sqrt{2}]\)

Check out this article on Limits and Continuity.

Integration of Tanx Summary

The integral of tan(x) means we are trying to find a function whose derivative is tan(x). In calculus, this process is called finding the antiderivative or indefinite integral.

We know that:

tan(x) = sin(x)/cos(x)

To integrate tan(x), we use a method called substitution. If we let u = cos(x), then du = -sin(x) dx. This helps us change the integral into a simpler form.

Now,

∫tan(x) dx = ∫sin(x)/cos(x) dx

= -∫du/u

= -ln|u| + C

Since u = cos(x), we replace it back:

∫tan(x) dx = -ln|cos(x)| + C

This is the most commonly used form. You can also write it as:

∫tan(x) dx = ln|sec(x)| + C

because -ln|cos(x)| is the same as ln|sec(x)|.

Here, ln is the natural logarithm and C is the constant of integration added to all indefinite integrals.

In summary, integrating tan(x) gives us a logarithmic function based on the cosine or secant of x.

Properties of the Integral of tan(x):

1. Derivative Relationship
   Since the derivative of -ln|cos(x)| is tan(x),
   we get:
   ∫tan(x) dx = -ln|cos(x)| + C
   (This can also be written as ln|sec(x)| + C)
2. Linearity
   If a is a constant, then:
   ∫a · tan(x) dx = a · ∫tan(x) dx
   = -a · ln|cos(x)| + C
3. Odd Function Property
   Since tan(x) is an odd function:
   ∫ from -a to a of tan(x) dx = 0
   (This holds only if the interval [-a, a] avoids the vertical asymptotes at x = π/2 + nπ)
4. Periodic Nature
   Since tan(x) has a period of π:
   ∫ from x to x+π of tan(t) dt = 0
   (Again, this holds only if the interval avoids points of discontinuity)
5. Integration in Definite Form
   When evaluating a definite integral like ∫ from a to b of tan(x) dx,
   make sure that cos(x) ≠ 0 in the interval [a, b] —
   because the integral becomes undefined at vertical asymptotes of tan(x),
   which occur at x = π/2 + nπ.

Examples of Integration of Tanx

Now that we have learnt the derivation and proof of integral of tanx let’s see some solved examples on integration of secx tanx.

Solved Example 1: Find the \({\int}tan^3(x)dx\)

Solution:

\({\int}tan^3(x)dx = {\int}tanx\cdot{tan^2x}dx\)

\({\int}tan^3(x)dx = {\int}tanx\cdot{tan^2x}dx\)

\({\int}tan^3(x)dx = {\int}tanx\cdot(sec^2x-1)dx\)

\({\int}tan^3(x)dx = {\int}(tanxsec^2x-tanx)dx\)

We know that, 

\({\int}tan^3(x)dx = {\int}tanxsec^2xdx-{\int}tanxdx\)

We will divide the problem into two integrations. 

\({\int}tanxsec^2xdx\)

We know that \(\frac{d}{dx}tanx=sec^2x\)

Let’s put \(tanx = u\) thus \(sec^2xdx = du\)

Thus, \({\int}tanxsec^2xdx\)=\({\int}udu\)

\({\int}tanxsec^2xdx\)=\(\frac{u^2}{2}\)

Now substituting \(u = tanx\) 

\({\int}tanxsec^2xdx\)=\(\frac{tanx^2}{2}\)

Now we take the main question:

\({\int}tan^3(x)dx = \frac{tanx^2}{2}-{\int}tanxdx\)

\({\int}tan^3(x)dx = \frac{tanx^2}{2}-log(sec(x))+c\)

Solved Example 2:Find ∫ tan⁵(x)·sec⁷(x) dx

Solution:
We rewrite the expression as:
∫ tan⁵(x)·sec⁷(x) dx = ∫ tan⁴(x)·sec⁶(x)·sec(x)·tan(x) dx

Now use the identity:
tan²(x) = sec²(x) − 1
⇒ tan⁴(x) = (tan²(x))² = (sec²(x) − 1)²

So the integral becomes:
∫ (sec²(x) − 1)²·sec⁶(x)·sec(x)·tan(x) dx
= ∫ (sec²(x) − 1)²·sec⁷(x)·tan(x) dx

Now substitute:
Let u = sec(x) ⇒ du = sec(x)·tan(x) dx

So:
∫ (u² − 1)² · u⁶ du
= ∫ (u⁴ − 2u² + 1) · u⁶ du
= ∫ (u¹⁰ − 2u⁸ + u⁶) du
= u¹¹/11 − 2u⁹/9 + u⁷/7 + C

Substituting back u = sec(x):
Final Answer:
∫ tan⁵(x)·sec⁷(x) dx = (sec¹¹(x))/11 − 2(sec⁹(x))/9 + (sec⁷(x))/7 + C

Solved Example 3:Find ∫ tan⁶(x)·sec⁴(x) dx

Solution:
We use the identity:
sec⁴(x) = (1 + tan²(x))·sec²(x)

So:
∫ tan⁶(x)·sec⁴(x) dx = ∫ tan⁶(x)·(1 + tan²(x))·sec²(x) dx
= ∫ (tan⁶(x) + tan⁸(x))·sec²(x) dx

Now let u = tan(x) ⇒ du = sec²(x) dx

So the integral becomes:
∫ (u⁶ + u⁸) du = u⁷/7 + u⁹/9 + C

Substitute back u = tan(x):
Final Answer:
∫ tan⁶(x)·sec⁴(x) dx = (tan⁷(x))/7 + (tan⁹(x))/9 + C

Solved Example 4: Find ∫ tan²(x) · sec⁴(x) dx

Solution:

Step 1: Use the identity
sec⁴(x) = (1 + tan²(x)) · sec²(x)

So the given integral becomes:
∫ tan²(x) · sec⁴(x) dx
= ∫ tan²(x) · (1 + tan²(x)) · sec²(x) dx
= ∫ (tan²(x) + tan⁴(x)) · sec²(x) dx

Step 2: Let
u = tan(x)
Then,
du = sec²(x) dx

Now the integral becomes:
∫ (u² + u⁴) du
= u³ / 3 + u⁵ / 5 + C

Step 3: Substitute back u = tan(x)

Final Answer:
∫ tan²(x) · sec⁴(x) dx = (tan³(x))/3 + (tan⁵(x))/5 + C

Hope this article on the Integral of Tanx was informative. Similarly, we can find limits of inverse trigonometric functions. Get some practice of the same on our free Testbook App. Download Now!