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  "body": "\n## Proof\n\nWe use the antiderivative $-\\cos$ of $\\sin$.\n\n1. Let $\\sin$ and $\\cos$ be the [[sine and cosine|0WB57D9488]], and let $F(x) = -\\cos x$.\n2. By the [[derivative of cosine|0V2DV29EGT]] and the [[constant multiple rule|0Z9JB71BS0]], $F$ is differentiable on $\\R$ and $F'(x) = -(-\\sin x) = \\sin x$.\n3. By the [[derivative of sine|00R4MWW532]], $\\sin$ is differentiable on $\\R$. [[Differentiable functions are continuous|07ZMX4ANJS]], so $\\sin$ and $F$ are continuous on $\\R$.\n4. Assume $a \u003c b$. By step 3, $\\sin$ is [[Riemann integrable|06SESV7GQY]] on $[a, b]$. The [[second fundamental theorem of calculus|0XYAFCN0PT]] applies by steps 2 and 3. It gives $\\int_a^b \\sin x \\, dx = F(b) - F(a) = \\cos a - \\cos b$.\n5. Assume $a = b$. By the convention for the [[definite integral|021WEM680W]], the integral is $0 = \\cos a - \\cos a$.\n6. Assume $a \u003e b$. By the same convention and step 4 on $[b, a]$, $\\int_a^b \\sin x \\, dx = -(\\cos b - \\cos a) = \\cos a - \\cos b$.\n7. So $\\int_a^b \\sin x \\, dx = \\cos a - \\cos b$ for all real $a$ and $b$.\n\n## Idea\n\nThe derivative of $\\cos$ is $-\\sin$, so $-\\cos$ is an antiderivative of $\\sin$. The fundamental theorem of calculus turns the integral into $-\\cos b - (-\\cos a)$.\n",
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