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diff --git a/site/content/integrating_sin2x.typ b/site/content/integrating_sin2x.typ
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-#import "/common.typ": post, link-self
-
-#show: post.with(
-    title: "Integrating sin² x",
-    date: datetime(year: 2018, month: 5, day: 1),
-)
-
-$
-  integral sin^2 x dif x = (x - sin x cos x) / 2 + C
-$
-
-= Using the Pythagorean identity
-
-Integrate by parts using $u = sin x, d u = cos x$ and $v = -cos x, d v = sin x$:
-
-$
-  & integral sin x sin x dif x = -sin x cos x - integral -cos x cos x dif x \
-  & integral sin^2 x dif x = -sin x cos x + integral cos^2 x dif x
-$
-
-Add $integral (sin x)^2 dif x$ to both sides:
-
-$
-  2 integral sin^2 x dif x = -sin x cos x + integral cos^2 x dif x + integral sin^2 x dif x
-$
-
-Apply integral linearity:
-
-$
-  2 integral sin^2 x dif x = -sin x cos x + integral (cos^2 x + sin^2 x) dif x
-$
-
-Apply the Pythagorean identity and divide both sides by 2:
-
-$
-  2 integral sin^2 x dif x &= -sin x cos x + integral 1 dif x \
-  &= -sin x cos x + x + C \
-  integral sin^2 x dif x &= (x - sin x cos x) / 2 + C
-$
-
-= Using the double angle formula
-
-#html.elem("aside", attrs: (class: "warning"))[
-  Careful! This derivation differs from the one in #link-self("integrating_cos2x")[integrating_cos2x] by just a sign.
-]
-Rewrite the integral using the double angle formula for the cosine:
-
-$
-  integral sin^2 x dif x = integral (1 - cos 2 x) / 2 dif x
-$
-
-Substitute $u = 2 x, d u = 2 dif x$:
-
-$
-  integral (1 - cos 2 x) / 2 dif x &= integral (1 - cos u) / 2 1 / 2 dif u \
-  &= 1 / 4 (integral 1 dif u - integral cos u dif u) \
-  &= (u - sin u + C) / 4
-$
-
-Substitute back and simplify:
-
-$
-  (u - sin u + C) / 4 &= (2 x - sin 2 x + C) / 4 \
-  &= (x - (sin 2 x) / 2) / 2 + C \
-  &= (x - sin x cos x) / 2 + C
-$