INTEGRATION of sinx/root(36-cos^2x)?
回答 (1)
2018-03-02 7:43 pm
✔ 最佳答案
Let u = (1/6)cosxThen, du = -(1/6)sinx dx and cosx = 6u
∫ {sinx dx / √(36 - cos²x)}
=(-6) ∫ {(-1/6) sinx dx / √[36 - (6u)²]}
=(-6) ∫ {du / √[6² - (6u)²]}
=(-6) ∫ {du / 6√(1 - u)²]}
= -∫{du/ √(1 - u)²]}
= -arccosu + C
= -arccos(cosx/6) + C
收錄日期: 2021-05-01 14:10:00
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https://hk.answers.yahoo.com/question/index?qid=20180302112123AAu9akk