Question #e1803

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Question #e1803

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Answer 1 · 2017-11-09T13:36:15

$\int sin 3 x d x = - \frac{1}{3} cos 3 x + C$ $intsin3xdx=-1/3cos3x+C$

Explanation:

by the chain rule

$\frac{d}{d x} (cos 3 x) = - 3 sin 3 x$ $d/(dx)(cos3x)=-3sin3x$

and as integration is the reverse of differentiation

$\int sin 3 x d x = - \frac{1}{3} cos 3 x + C$ $intsin3xdx=-1/3cos3x+C$

Answer 2 · 2017-11-09T13:49:09

$- (\frac{1}{3}) cos 3 x + C$ $-(1/3)cos3x + C$

Explanation:

Since we are dealing with integration, we must always include a "+C" as the derivative of any Constant is 0

Questions to ask:

The derivative of what is $sin x$ $sinx$ ?

Well, if you know your trig rules, we know that the derivative of $cos x$ $cosx$ is $- sin x$ $-sinx$ , so we can conclude that the derivative of $- cos x$ $-cosx$ is $sin x$ $sinx$ .

What about the $3 x$ $3x$ ?

Well, since there is a 3x# inside the cos , we can infer that the chain rule is involved:

Ex.

$\frac{d}{d x} sin 4 x = (cos 4 x) (4)$ $d/dx sin4x = (cos4x)(4)$

$\frac{d}{d x} cos 6 x = - (sin 6 x) (6)$ $d/dx cos6x = -(sin6x)(6)$

So, knowing the chain rule, we know that when this was derived, the $3 x$ $3x$ was derived and multiplied by the entire equation, So we must ask:

If the $3$ $3$ is multiplied by the equation, then why is there a $1$ $1$ in front instead of a $3$ $3$ ?

For this to be true there must have been a number already in front of the $- cos 3 x$ $-cos3x$ so:

$3 (?) = 1$ $3(?) = 1$

$3 (\frac{1}{3}) = 1$ $3(1/3) = 1$

So now we know that, originally, there was a (1/3) in front of the equation, so we conclude that:

$\int sin (3 x) d x = - (\frac{1}{3}) cos 3 x + C$ $intsin(3x)dx = -(1/3)cos3x + C$

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Question #e1803

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