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JulijaS [17]

Answer:

$\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty)$

General Formulas and Concepts:

Pre-Calculus

Calculus

Limits
Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_{x \to c} x = c$
Integrals
Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$
Trig Integration
Improper Integrals

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^0_\infty {cos(x)} \, dx$

Step 2: Integrate

[Improper Integral] Rewrite: $\displaystyle \lim_{a \to \infty} \int\limits^0_a {cos(x)} \, dx$
[Integral] Trig Integration: $\displaystyle \lim_{a \to \infty} sin(x) \bigg| \limits^0_a$
[Integral] Evaluate [Integration Rule - FTC 1]: $\displaystyle \lim_{a \to \infty} sin(0) - sin(a)$
Evaluate trig: $\displaystyle \lim_{a \to \infty} -sin(a)$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle -sin(\infty)$

Since we are dealing with infinity of functions, we can do a numerous amount of things:

Since -sin(x) is a shift from the parent graph sin(x), we can say that -sin(∞) = sin(∞) since sin(x) is an oscillating graph. The values of -sin(x) already have values in sin(x).
Since sin(x) is an oscillating graph, we can also say that the integral actually equates to undefined, since it will never reach 1 certain value.

∴ $\displaystyle \int\limits^0_\infty {cos(x)} \, dx = sin(\infty) \ or \ \text{unde}\text{fined}$