Question #5454a

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Question #5454a

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Answer 1 · 2017-11-14T18:33:57

$\int sin 6 x d x = - \frac{1}{6} cos 6 x + C$ $intsin6xdx=-1/6cos6x+C$

Explanation:

$\int sin 6 x d x$ $intsin6xdx$

Let $u = 6 x$ $u=6x$
$\frac{d u}{d x} = 6$ $(du)/dx=6$
$d x = \frac{d u}{6}$ $dx=(du)/6$

$\int sin 6 x d x = \int sin u \frac{d u}{6} = \frac{1}{6} \int sin u d u = \frac{1}{6} (- cos u + C) = - \frac{1}{6} cos 6 x + C$ $intsin6xdx=intsinu(du)/6=1/6intsinudu=1/6(-cosu+C)=-1/6cos6x+C$

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Question #5454a

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