Question

Evaluate the integral. (Use C for the constant of integration.)
∫ 5tan³(x)sec(x)dx

Answer 1

The integral of 5tan³(x)sec(x)dx is evaluated by using u-substitution where u = tan(x). The integral then becomes 5u² du, which is integrated to get ⅓u³. Substituting back for u, the result is ⅓tan³(x) + C.

Step-by-step explanation:

The question asks to evaluate the integral of the function 5tan³(x)sec(x)dx. To solve this, we can recognize that tan(x)sec(x)dx is the derivative of tan²(x), which suggests a substitution.

Step-by-step Explanation:

Let u = tan(x). Then, du = sec²(x)dx.

Since we have tan³(x), we rewrite it as (tan(x))²tan(x) = u²u, and sec(x)dx as du/sec²(x). This gives us the integral 5u² du.

1. Perform the integral of 5u² du to get ⅓u³.
2. Substitute back tan(x) for u to get the antiderivative, which is ⅓tan³(x).
3. Add the constant of integration C.

To evaluate the integral ∫ 5tan³(x)sec(x)dx, we can use the trigonometric identity: tan²(x) = sec²(x) - 1. Let's rewrite the integral in terms of tan(x) only.

∫ 5tan³(x)sec(x)dx = ∫ 5tan²(x)(sec(x))tan(x)dx = ∫ 5(sec²(x) - 1)(sec(x))tan(x)dx

Now, let's make a substitution: let u = sec(x), then du = sec(x)tan(x)dx. The integral becomes:

∫ 5(u² - 1)du = ∫ 5u³ - 5u du

Integrating, we get:

½u⁴ - ½u² + C = ½(sec²(x))² - ½sec²(x) + C = ½sec⁴(x) - ½sec²(x) + C, where C is the constant of integration.

Therefore, the integral of 5tan³(x)sec(x)dx is ⅓tan³(x) + C.