Evaluation of the Integral of Tan(u) du Using the Substitution Method

∫tan u du

To evaluate the integral of tan(u) du, we can use the substitution method

To evaluate the integral of tan(u) du, we can use the substitution method.

Let’s set u = tan(x), which means du = sec^2(x) dx. Rearranging this, we have dx = du / sec^2(x).

Now substitute these values into the integral:

∫tan(u) du = ∫tan(x) (du / sec^2(x))

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

∫tan(u) du = ∫tan(x) (du / (1 + tan^2(x)))

Next, notice that the numerator and denominator both have tan(x). We can perform another substitution, let’s call it v.

v = tan(x), so dv = sec^2(x) dx. Rearranging, we have dx = dv / sec^2(x).

Substituting again, we get:

∫tan(x) (du / (1 + tan^2(x))) = ∫v (dv / (1 + v^2))

Now, the integral becomes:

∫v / (1 + v^2) dv

To evaluate this integral, we can use a u-substitution. Let’s set u = 1 + v^2, so du = 2v dv.

Rearranging this, we have dv = du / (2v).

Substituting these values into the integral, we get:

∫v / (1 + v^2) dv = ∫(u – 1) / (2u) du

Simplifying the numerator, we have:

∫(u – 1) / (2u) du = ∫(u / 2u) – (1 / 2u) du

This further simplifies to:

∫1/2 – 1/(2u) du

Taking the integral of each term separately gives:

(1/2) ∫1 du – (1/2) ∫1/u du

∫1 du is simply u, so the first term becomes:

(1/2) u

∫1/u du is ln|u|, so the second term becomes:

-(1/2) ln|u|

Therefore, the final integral is:

(1/2) u – (1/2) ln|u| + C

Substituting back for u, we have:

(1/2) tan(x) – (1/2) ln|tan(x)| + C

Therefore, the integral of tan(u) du is (1/2) tan(x) – (1/2) ln|tan(x)| + C, where C is the constant of integration.

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