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  "body": "\n## Statement\n\nLet $g \\colon [a, b] \\to \\mathbb{R}$ be [[continuously differentiable|06SKCHDQT5]], and let $f$ be continuous on an interval that contains $g([a, b])$.\nThen\n$$\\int_a^b f(g(x)) \\, g'(x) \\, dx = \\int_{g(a)}^{g(b)} f(u) \\, du.$$\n\n## Remarks\n\nLet $F$ be an [[antiderivative|0NC2A3HWGT]] of $f$. By the [[chain rule|0RXX2PT1MQ]], $F \\circ g$ is an antiderivative of $(f \\circ g) \\, g'$. Both sides equal $F(g(b)) - F(g(a))$.\nThe function $g$ need not be injective.\nFor example, with $u = x^2$, $\\int_0^1 2x \\cos(x^2) \\, dx = \\int_0^1 \\cos u \\, du = \\sin 1$.\nMathlib states it as `intervalIntegral.integral_comp_mul_deriv`.\n\n## Explanations\n\n### L1 human\n\nSubstitution is the chain rule read backwards.\nWe replace an expression $g(x)$ by a new variable $u$, and $g'(x) \\, dx$ by $du$.\n",
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