5.4 Indefinite Integrals and the Net Change Theorem/3: Difference between revisions

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& \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} \\[2ex]
& \frac{d}{dx} {[\sin{x} - \frac{1}{3} \sin^3{x} +C]} \\[2ex]


& ={\cos{x} - \frac{1}{3}\cdot 3\sin{x^2} \cos{x} +0} \\[2ex]
& ={\cos{x} - \frac{1}{3}\cdot 3\sin^2{x} \cos{x} +0} \\[2ex]


& =\cos{x} - \sin^2{x}\cos{x} \\[2ex]
& =\cos{x} - \sin^2{x}\cos{x} \\[2ex]

Revision as of 17:11, 13 September 2022