---
tag: 1R7Z7BT67Q
kind: proof
name: Proof of the integral of cosine
msc: [26A36]
proves: 04WTBQ7APQ
uses: [0WB57D9488, 00R4MWW532, 0V2DV29EGT, 07ZMX4ANJS, 06SESV7GQY, 0XYAFCN0PT, 021WEM680W]
sources:
  - url: https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
    locator: Second part
---

## Proof

We use the antiderivative $\sin$ of $\cos$.

1. Let $\sin$ and $\cos$ be the [[sine and cosine|0WB57D9488]].
2. By the [[derivative of sine|00R4MWW532]], $\sin$ is differentiable on $\R$ and $\sin' = \cos$.
3. By the [[derivative of cosine|0V2DV29EGT]], $\cos$ is differentiable on $\R$. [[Differentiable functions are continuous|07ZMX4ANJS]], so $\sin$ and $\cos$ are continuous on $\R$.
4. Assume $a < b$. By step 3, $\cos$ is [[Riemann integrable|06SESV7GQY]] on $[a, b]$. The [[second fundamental theorem of calculus|0XYAFCN0PT]] with $F = \sin$ applies by steps 2 and 3. It gives $\int_a^b \cos x \, dx = \sin b - \sin a$.
5. Assume $a = b$. By the convention for the [[definite integral|021WEM680W]], the integral is $0 = \sin a - \sin a$.
6. Assume $a > b$. By the same convention and step 4 on $[b, a]$, $\int_a^b \cos x \, dx = -(\sin a - \sin b) = \sin b - \sin a$.
7. So $\int_a^b \cos x \, dx = \sin b - \sin a$ for all real $a$ and $b$.

## Idea

The derivative of $\sin$ is $\cos$, so $\sin$ is an antiderivative of $\cos$. The fundamental theorem of calculus turns the integral into the difference $\sin b - \sin a$.