// SPDX-License-Identifier: CC-BY-SA-4.0
// Copyright © Paweł Redman

#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("math/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
$