#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
$