How do you find the antiderivative of $cos (5 x)$ $cos(5x)$ ?

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How do you find the antiderivative of $cos (5 x)$ $cos(5x)$ ?

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Answer 1 · 2016-10-28T21:25:23

Say that:

$y = sin (k x)$ $y=sin(kx)$ whereby k is a constant.

Now, transform this into:

$y = sin (u)$ $y=sin(u)$ whereby $u = k x$ $u=kx$ .

If this is the case:

$\frac{d y}{d u} = cos (u) = cos (k x)$ $(dy)/(du)=cos(u)=cos(kx)$

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

And this means that:

$\frac{d y}{d u} \cdot \frac{d u}{d x} = \frac{d y}{d x} = k cos (k x)$ $(dy)/(du)*(du)/(dx)=(dy)/(dx)=kcos(kx)$

Now, when k=5:

When $y = sin (5 x)$ $y=sin(5x)$ , $\frac{d y}{d x} = 5 cos (5 x)$ $(dy)/(dx)=5cos(5x)$ .

Ok... So what happens when:

$y = \frac{1}{5} sin (5 x)$ $y=1/5sin(5x)$ ??

Well...

$5 y = sin (5 x)$ $5y=sin(5x)$

Differentiating both sides of the equation and remembering that $\frac{d y}{d y} \cdot \frac{d y}{d x} = \frac{d y}{d x}$ $(dy)/(dy)*(dy)/(dx)=(dy)/(dx)$ you get...

$5 \cdot \frac{d y}{d x} = 5 cos (5 x)$ $5*(dy)/(dx)=5cos(5x)$

Then dividing both sides of this equation by 5 you get...

$\frac{d y}{d x} = cos (5 x)$ $(dy)/(dx)=cos(5x)$

This means that:

$\int cos (5 x) d x = \frac{1}{5} sin (5 x) + C$ $intcos(5x)dx=1/5sin(5x)+C$

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How do you find the antiderivative of $cos (5 x)$ $cos(5x)$ ?

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