Integration of Tan x

Integration of tan x is equivalent to log |sec x| + C.

Integration of Tan x means finding the integral of the trigonometric function tan x. The integral of tan x with respect to x can be written as ∫ tan x dx. Here, we need to find the indefinite integral of tan x. So, the integration of tan x results in a new function and an arbitrary constant C. Let’s know the formula for the integration of tan x, along with the derivation and examples.

Integration of Tan x Formula

The formula for the integration of tan x into dx is given by:

Or

Integration of Tan x dx Derivation

∫ tan x dx

We know that tan A = sin A/cos A

Thus, ∫ tan x dx = ∫ (sin x /cos x) dx

= ∫ (1/cos x) sin x dx

Let’s apply the substitution method of integration.

Let t = cos x

⇒ dt = – sin x dx

⇒ sin x dx = -dt

So, ∫ tan x dx = ∫(sin x /cos x) dx = ∫ (1/t) (-dt) = – ∫ (1/t) dt

– log |t| + C

= -log |cos x| + C

= log |(cos x)^-1| + C

= log |1/cos x| + C

= log |sec x| + C

Therefore, ∫ tan x dx = -log |cos x| + C = log |sec x| + C

Examples on Integration of Tan x

Example 1: Integration of Tan x whole square

Integral of tan x whole square can be written as:

∫ (tan x)²

Let us find the integral of (tan x)² with respect to dx.

∫ (tan x)² dx = ∫ tan²x dx

Using the identity sec²A – tan²A = 1,

∫ tan²x dx = ∫ (sec²x – 1) dx

= ∫ sec²x dx – ∫ 1 dx

Using the standard integration formulas, ∫sec²x dx = tan x + C and ∫ 1 dx = x + C, we get;

= tan x – x + C

Therefore, ∫ (tan x)² dx = (tan x) – x + C

Example 2: Integration of under root tan x

Integral of under root tan x can be written as:

∫ √(tan x)

Let’s find the integral of under root tan x with respect to x.

∫ √(tan x) dx

Let tan x = t² such that tan²x = t⁴ and sec²x = 1 + t⁴.

Differentiate with respect to x.

⇒ sec²x dx = 2t dt

⇒ dx = [2t / (1 + t⁴)]dt

Dividing the numerator and denominator of RHS by t², we get;

Let us substitute t + 1/t = m and t – 1/t = z.

Also, by differentiating these equations with respect to t, we get;

1 – 1/t² = dm/dt and 1 + 1/t² = dz/dt

⇒ (1 – 1/t²) dt = dm and (1 + 1/t²) dt = dz

Now, substitute t = √(tan x) and 1/t = 1/√(tan x) = √(cot x).

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