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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("math/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
$
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