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+#import "/common.typ": post, link-self
+
+#show: post.with(
+    title: "Integrating cos² x",
+    date: datetime(year: 2018, month: 5, day: 1),
+)
+
+$
+  integral cos^2 x dif x = (x + sin x cos x) / 2 + C
+$
+
+= Using the Pythagorean identity
+
+Integrate by parts using $u = cos x, d u = -sin x$ and $v = sin x, d v = cos x$:
+
+$
+  & integral cos x cos x dif x = sin x cos x - integral -sin x sin x dif x \
+  & integral cos^2 x dif x = sin x cos x + integral sin^2 x dif x
+$
+
+Add $integral (cos x)^2 dif x$ to both sides:
+
+$
+  2 integral cos^2 x dif x = sin x cos x + integral sin^2 x dif x + integral cos^2 x dif x
+$
+
+Apply integral linearity:
+
+$
+  2 integral cos^2 x dif x = sin x cos x + integral (sin^2 x + cos^2 x) dif x
+$
+
+Apply the Pythagorean identity and divide both sides by 2:
+
+$
+  2 integral cos^2 x dif x &= sin x cos x + integral 1 dif x \
+  & = sin x cos x + x + C \
+  integral cos^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_sin2x")[integrating_sin2x] by just a sign.
+]
+
+Rewrite the integral using the double angle formula for the cosine:
+
+$
+  integral cos^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
+$