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

Oct 28, 2016

Say that:

$y = \sin \left(k x\right)$ whereby k is a constant.

Now, transform this into:

$y = \sin \left(u\right)$ whereby $u = k x$.

If this is the case:

$\frac{\mathrm{dy}}{\mathrm{du}} = \cos \left(u\right) = \cos \left(k x\right)$

$\frac{\mathrm{du}}{\mathrm{dx}} = k$

And this means that:

$\frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}} = k \cos \left(k x\right)$

Now, when k=5:

When $y = \sin \left(5 x\right)$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 5 \cos \left(5 x\right)$.

Ok... So what happens when:

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

Well...

$5 y = \sin \left(5 x\right)$

Differentiating both sides of the equation and remembering that $\frac{\mathrm{dy}}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dx}}$ you get...

$5 \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 5 \cos \left(5 x\right)$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \cos \left(5 x\right)$

This means that:

$\int \cos \left(5 x\right) \mathrm{dx} = \frac{1}{5} \sin \left(5 x\right) + C$