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| author | Paweł Redman <pawel.redman@gmail.com> | 2026-06-14 19:11:45 +0200 |
|---|---|---|
| committer | Paweł Redman <pawel.redman@gmail.com> | 2026-06-14 19:11:45 +0200 |
| commit | cbd693043b83c1d997c52f6c3fca83c6a7a37e28 (patch) | |
| tree | 2dc74ba212b077857f2da37fe99960631690ffd6 /site/content/integrating_sin2x.typ | |
| parent | 0c1c00b3fc2a880c18f6f98d87372dc91451b0b0 (diff) | |
Migrate math notes to Typst.
Co-authored-by: Cursor <cursoragent@cursor.com>
Diffstat (limited to 'site/content/integrating_sin2x.typ')
| -rw-r--r-- | site/content/integrating_sin2x.typ | 66 |
1 files changed, 0 insertions, 66 deletions
diff --git a/site/content/integrating_sin2x.typ b/site/content/integrating_sin2x.typ deleted file mode 100644 index 65b17a6..0000000 --- a/site/content/integrating_sin2x.typ +++ /dev/null @@ -1,66 +0,0 @@ -#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 -$ |
