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C65-12. {$$\color{blue}{ y = \ln {t \over {\sqrt{1-t^2}}}} \text{ and } \color{red}{y^\prime}$$}

Blackboard Area

C98-17. {$$\color{blue}{ y = \sin^3 }$$}

{$$\color{red}{y= -\cos x + {1 \over 3}\cos^3 }$$}{$$\color{green}{y= -\cos x + {1 \over 3}\cos^3 +2}$$}{$$\color{maroon}{y= -\cos x + {1 \over 3}\cos^3 -2}$$}

C98-17. {$\displaystyle \int \sin^3 x dx$}

{$u=\cos x,\;du= - \sin x,\;{{-du} \over { \sin x}}=dx$}

{\begin{align} &\int \sin^3 x dx = \int \sin x(1- cos^2x) dx \cr &= \int \sin x(1 - u^2){{-1} \over { \sin x}}du \cr &= \int -1 + u^2du \cr &= -\int 1du + \int u^2du \cr &= -u + {1 \over 3} u^3 \cr &= -\cos x + {1 \over 3} \cos^3x \end{align}}

Check:

{\begin{align} & {d \over {du}}(-\cos x + {1 \over 3} \cos^3x) = -(-\sin x) + 3({1 \over 3})\cos^2 x(-\sin x) \cr &= \sin x - \sin x\cos^2 x = \sin x(1-\cos^2 x) \cr &= \sin{x} \sin^2 x = \sin^3 x \end{align}}

Notes:

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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March 18, 2018

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