## ∫sec²udu

### To evaluate the integral ∫sec²(u) du, we can use the power rule for integrals

To evaluate the integral ∫sec²(u) du, we can use the power rule for integrals. The power rule states that ∫x^n dx = (1/(n+1)) * x^(n+1) + C, where C is the constant of integration.

In this case, the power of sec²(u) is 2, so we can rewrite the integral as ∫sec²(u) du = (1/3) * sec²(u)³ + C.

However, when dealing with trigonometric functions, it is often more convenient to use trigonometric identities to simplify the integral.

The trigonometric identity that relates sec(x) with tan(x) is sec²(x) = 1 + tan²(x). By rearranging this identity, we can solve for sec²(x) as sec²(x) = tan²(x) + 1.

So, in the integral ∫sec²(u) du, we can substitute sec²(u) with tan²(u) + 1:

∫sec²(u) du = ∫(tan²(u) + 1) du.

Now we can split the integral into two parts:

∫tan²(u) du + ∫1 du.

The first integral, ∫tan²(u) du, can be solved by using a trigonometric identity that relates tan(x) with sec²(x). The identity is: sec²(x) = tan²(x) + 1. Rearranging the terms, we have tan²(x) = sec²(x) – 1.

Using this identity, the integral becomes:

∫tan²(u) du = ∫(sec²(u) – 1) du.

Now we can substitute the first part of the integral with sec²(u) – 1:

∫tan²(u) du = ∫(sec²(u) – 1) du = ∫sec²(u) du – ∫du.

The first integral, ∫sec²(u) du, is what we initially started with and its value is (1/3) * sec²(u)³ + C.

The second integral, ∫1 du, is a simple integral which evaluates to u + C.

Combining both results, we obtain the final evaluation of the integral:

∫sec²(u) du = (1/3) * sec²(u)³ + u + C.

Therefore, the solution to the integral ∫sec²(u) du is (1/3) * sec²(u)³ + u + C, where C is the constant of integration.

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