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  "body": "\n## Proof\n\nWe use the antiderivative $\\sin$ of $\\cos$.\n\n1. Let $\\sin$ and $\\cos$ be the [[sine and cosine|0WB57D9488]].\n2. By the [[derivative of sine|00R4MWW532]], $\\sin$ is differentiable on $\\R$ and $\\sin' = \\cos$.\n3. By the [[derivative of cosine|0V2DV29EGT]], $\\cos$ is differentiable on $\\R$. [[Differentiable functions are continuous|07ZMX4ANJS]], so $\\sin$ and $\\cos$ are continuous on $\\R$.\n4. Assume $a \u003c b$. 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 $\\int_a^b \\cos x \\, dx = \\sin b - \\sin a$.\n5. Assume $a = b$. By the convention for the [[definite integral|021WEM680W]], the integral is $0 = \\sin a - \\sin a$.\n6. Assume $a \u003e b$. By the same convention and step 4 on $[b, a]$, $\\int_a^b \\cos x \\, dx = -(\\sin a - \\sin b) = \\sin b - \\sin a$.\n7. So $\\int_a^b \\cos x \\, dx = \\sin b - \\sin a$ for all real $a$ and $b$.\n\n## Idea\n\nThe derivative of $\\sin$ is $\\cos$, so $\\sin$ is an antiderivative of $\\cos$. The fundamental theorem of calculus turns the integral into the difference $\\sin b - \\sin a$.\n",
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