Solving the Integral of tan(x) using Substitution and Partial Fractions

∫tanxdx

To evaluate the integral ∫tan(x)dx, we can use a technique called substitution

To evaluate the integral ∫tan(x)dx, we can use a technique called substitution. Here’s how:

Let u = tan(x), then du = sec^2(x)dx.
Rearranging the equation, we get dx = du / sec^2(x).

Substituting these expressions into the integral, we have:
∫tan(x)dx = ∫u * (du / sec^2(x))

Using the identity sec^2(x) = 1 + tan^2(x), we can rewrite the integral as:
∫u * (du / (1 + u^2))

Now, we can use a standard integration technique for rational functions: decomposing the expression into partial fractions.

Let’s find the partial fraction decomposition for 1 / (1 + u^2):
1 / (1 + u^2) = A / (1 + u) + B / (1 – u)

Multiplying both sides by (1 + u)(1 – u), we get:
1 = A(1 – u) + B(1 + u)

Expanding and comparing coefficients, we find:
1 = (A + B) + (B – A)u

Equating the coefficients on both sides, we have:
A + B = 1 (coefficient of 1)
B – A = 0 (coefficient of u)

Solving this system of equations, we can determine that A = B = 1/2.

Now, we can rewrite the integral in terms of the partial fractions:
∫u * (du / (1 + u^2)) = ∫(1/2)/(1 + u) du + ∫(1/2)/(1 – u) du

Integrating each term separately, we get:
(1/2) * ln|1 + u| – (1/2) * ln|1 – u| + C

Substituting back u = tan(x), we have:
(1/2) * ln|1 + tan(x)| – (1/2) * ln|1 – tan(x)| + C

Thus, the integral of tan(x) is given by the expression:
(1/2) * ln|1 + tan(x)| – (1/2) * ln|1 – tan(x)| + C, where C is the constant of integration.

∫tanxdx