How to Integrate the Tangent Function | Step-by-Step Guide for ∫tan(x) dx

int tanx

The expression “int tanx” refers to finding the integral of the tangent function, ∫tan(x) dx

The expression “int tanx” refers to finding the integral of the tangent function, ∫tan(x) dx.

To integrate this function, we can use a technique called integration by substitution. Here’s a detailed step-by-step process:

1. Start by using a substitution where you let u = tan(x).
This means that du = sec^2(x) dx.

2. Rewriting the integral using the substitution, we have:
∫tan(x) dx = ∫(u)(du/sec^2(x))

3. Now, we need to get rid of the sec^2(x) term. We know that sec^2(x) is equal to 1 + tan^2(x), so we can rewrite the integral as:
∫(u)(du/(1 + u^2))

4. We can use partial fractions to break down the expression into simpler fractions. Factoring the denominator, we have:
1 + u^2 = (1 + iu)(1 – iu)
Therefore, we can write:
∫(u)(du/(1 + u^2)) = ∫(A/(1 + iu) + B/(1 – iu))du

5. Expanding the right side, we have:
u = A(1 – iu) + B(1 + iu)
Re-arranging, we get:
u = (A + B) + (Bi – Ai)

Equating real and imaginary parts, we have:
A + B = 0 (for the real part)
Bi – Ai = u (for the imaginary part)

Solving these equations, we find that A = B = 1/2i.

6. Substituting the partial fractions back into the integral, we obtain:
∫(1/2i)(1/(1 + iu) + 1/(1 – iu))du

7. Now, we can integrate each term separately:
∫(1/2i)(1/(1 + iu) + 1/(1 – iu))du = (1/2i) * (ln(1 + iu) – ln(1 – iu)) + C
(where C is the constant of integration)

8. Simplifying the expression, we have:
(1/2i) * (ln(1 + iu) – ln(1 – iu)) + C
= (1/2i) * ln((1 + iu)/(1 – iu)) + C
= (1/2i) * ln((1 + i(tan(x)))/(1 – i(tan(x)))) + C

And that is the final answer for the integral of tan(x):
∫tan(x) dx = (1/2i) * ln((1 + i(tan(x)))/(1 – i(tan(x)))) + C

More Answers: