How do you integrate sin^2 (4x/5)?
回答 (3)
2016-05-01 11:52 am
cos 2Ө = 1 - 2 sin²Ө
sin²Ө = 1/2 - (1/2) cos 2Ө
sin²(4x/5) = 1/2 - (1/2) cos (8x/5)
∫ sin²(4x/5) dx = ∫ 1/2 - (1/2) cos (8x/5) dx
∫ sin²(4x/5) dx = x/2 - (5/16) sin (8x/5) + C
sin²Ө = 1/2 - (1/2) cos 2Ө
sin²(4x/5) = 1/2 - (1/2) cos (8x/5)
∫ sin²(4x/5) dx = ∫ 1/2 - (1/2) cos (8x/5) dx
∫ sin²(4x/5) dx = x/2 - (5/16) sin (8x/5) + C
2016-05-01 6:36 am
2016-05-01 6:35 am
You integrate it by first expressing it as a first power.
cos (2x) = cos^2 x - sin^2 x
and
sin^2 x + cos^2 x = 1
or cos^2 x = 1 - sin^2 x
so cos (2x) = 1 - 2 sin^2 x
and sin^2 x = (1/2) [1 - cos (2x)]
integral sin^2 (4x/5) dx
= integral (1/2) [1 - cos (8x/5)] dx
= (1/2) [x - (5/8) sin (8x/5)] + C
where C is an arbitrary constant
cos (2x) = cos^2 x - sin^2 x
and
sin^2 x + cos^2 x = 1
or cos^2 x = 1 - sin^2 x
so cos (2x) = 1 - 2 sin^2 x
and sin^2 x = (1/2) [1 - cos (2x)]
integral sin^2 (4x/5) dx
= integral (1/2) [1 - cos (8x/5)] dx
= (1/2) [x - (5/8) sin (8x/5)] + C
where C is an arbitrary constant
收錄日期: 2021-04-23 20:45:26
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https://hk.answers.yahoo.com/question/index?qid=20160430221457AA6tFF5