Evaluate: `int(cosx)/(sqrt(sin^2x-2sinx-3))dx`
`(cos x)/(sqrt(sin^(2)x-2sinx-3))`
integrate (sin x - cos x)/(sqrt(sin 2x)) dx
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積分を評価します。 sin^(2)x cos^(3)x dx
cosx/sin^2xの積分(解)
∫cos(πx)(sin(πx))^(2022π) dx [0, 1/2] = ?
∫(sin(2x) - sin²(x))/(cos(2x) - cos²(x)) dx. MIT Integration Bee 2018, Question 15, Qualifying Exam.
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